B   M   SOI   b5S 


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Descriptive  Geometry 


for  Students  in 

Engineering  Science 

and 

Architecture 


A  CAREFULLY  GRADED  COURSE  OF  INSTRUCTION 

BY 

HENRY  F.  ARMSTRONG 

Associate  Professor  of  Descriptive  Geometry  and  Drawing,  McGill  University 


FIRST  EDITION 

FIRST    THOUSAND 


4 


NEW    YORK: 
JOHN   WILEY   e^    SONS,  Inc. 

London:    CHAPMAN    i:    HALL,     Limited 


Copyright,  igiS 

BY 

HENRY   F.  ARMSTRONG 


THE  SCIENTIFIC  PRESS 

ROBERT   DRUMMOND  AND  COMPANY 

BROOKLYN,   N.  Y. 


PREFACE 


The  writer  offers  the  contents  of  this  Text  Book  as  the  result  of  over 

rfounrtht'^r"','"'""*^  "^  """"'p'^"  ^^-^^'^y-  ^-^  ^^  ^e 

has  found  that  a  logical  presentation  of  the  subject,  concerning  itself  largely 
with  the  best  sequence  possible,  and  with  the  introduction,  at  an  early 
stage,  of  practical  applications  and  well-graded  exercises  by  which  the  first 
rules  and  principles  can  be  practised,  is  far  more  preferable  and  satisfactory 
in  Its  results  than  the  mode  of  procedure  and  division  of  the  subject  matter 
commonly  adopted  in  other  text  books  on  the  subject 

Difficult  and  complex  phases  of  Descriptive  Geometry  and  problems 
beyond  what  the  average  student  has  time  to  assimilate  during  a  limited 
college  course  and  beyond  what  is  worth  while  in  such  a  course,  are  avoided 
because  ol  he  tendency  for  students  to  become  thereby  discouraged  and  to 
sX^LLtr^  ""  ^*"  ^'"""'  "'^^™'-  P™™  '°  '^  -  '"-esting 
No  pretense,  therefore,  has  been  made  to  place  before  students  an 
exhaustive  treatise  on  the  subject,  but  rather  to  deal  with  the  essenti" 
facts  and  methods  readily  arrived  at  and  their  useful  application.  To  this 
end,  the  student  is  frequently  required  to  work,  independently,  exercises  on 
the  problems  of  which  illustrations  are  given  and  discussed,  aiid  himse  to 
propose  questions  or  data  for  new  exercises  to  be  worked 

It  will  be  noticed  that  the  study  of  the  subject  is  conducted  in  such 
a  way  that  Part  I  of  the  Text  Book  may  serve  as  an  Introductory  Course 
more  or  less  complete  in  itself,  and  may  be  sufficient  and  suitable  for  a 
first  term  in  a  Science  Course  Curriculum,  or  for  the  upper  forms  in  High 
Schools;  or.  Parts  I  and  II  may  be  taken  in  the  First  Year  and  Part  HI 
VrJy  I""  '  ^"^■""'•^'  "^  ^°"^="  Course  in  Science,  if  in 
undertaken     ''  '    "   ""'    ^'"''"^'"^   '"""'"'"'    """=    '"'   ""^   "*°'»    '"    be 

so    liHle  ""^    ""''^'  •"'    f"''    '"''''''''    '^'    ™^Sinatixe    faculty    of    the    mind, 

0    little    exercised    in    the    ordinary    school    curriculum,    is    afforded    a    .ood 

traming.    while    the   practice    in   grasping   a    collection    of   detailed    conditions 

the  «ould-be  engineer  or  architect  in  his  future  career. 
Janu.lry,  11)15. 


CONTENTS 


SEC. 

1.  The  Projection  Planes 

2.  Lines  and  Their  Inclinations. 


PAGE 


4 


PART  ONE 

CHAPTER   I 

CH.VPTER  II 

3.  Plan  and  I^levation  Fo-ms  for  Plane  Rectilineal  Figures g 

4.  Plan  and  Elevation  Forms  for  the  Circle 10 

chapti:r  III 

5.  Oblique  Planes,  Their  Traces  and  Inclinations 13 

6.  Traces  of  Lines 17 

CHAPri:R  IV 

7.  Shadows  of  Lines ro 

8.  Shadows  of  Plane  Figures 22 

CHAPTER  \' 

().  The  '■  Com[)ound  Angle  "  for  Lines 25 

10.  Projections  of  Plane  Figures,  Involving  the  Compound  Angle 29 

CHAPTER    VI 

11.  Projeition  of  Simple  Solids 31 

I :;.  Shadows  oi  Solitls  and  of  Groups  of  Solids ;5 


IWR'I"    I'WO 

CHAPTER  \H 

I  ^  Rabattemcnt  of  Planes,  and  Projections  of  Figures  Involving  Its  LTse 40 

14.  Points  and  Lines  in  Relation  to  Ohliiiue  Planes 45 

15.  Intersection  of  Obli(iue  Planes  with   Each  Other,  and  of  Lines  with  Oblique  Planes 4S 


CONTENTS 


CHAPTER  VIII 


SEC. 


[6.  Parallel  Planes. 


and  Planes  Dividing  These  Angles.  .     56 


17'.  Dihedral  Angles  Contained  by  Oblique  Intersecting  Planes, 

CHAPTER  IX 

x8.  Projection  of  Rectilineal  Angles,  and  of  Figures  and  Solids  Involving  the  Same 6^ 

19.  The  Tetrahedron  and  the  Oclahedron 

CHAPTER  X 

69 

20.  Axometric  and  Isometric  Projection ^^ 

21.  Axometric  Projection,  Continued 

CHAPTER  XI 


22.  Sections  of  Simple  Solids... 

23.  Traces  of  Curved  Surfaces. 


PART  THREE 

CHAPTER   XII 


24.  Tangent  Planes  to  Cones  and  Cylinders ;.:' ^  "r^-"  ' ,  "  ^  '''' 

25.  Projection  of  Solids  dependent  on  Tangent  Planes  to  Right  Circular  Cones 


CHAPTER   XIII 

26.  Tangent  Planes  to  a  Sphere,  and  the  Finding  of  an  Oblique  Plane  with  Given  Inclinations 93 

27.  Tangent  Planes  Common  to  Three  Given  Spheres 

CHAPTER   XIV 

Surfaces m 

30.  Surfaces  of  Revolution,  and  the  Screw  Thread 


CHAPTER   XV 


31.  Radial  Projection.     Perspective  Projection . 

32.  Perspective  Projection,  continued 


c 


/- 


DESCRIPTIVE    GEOMETRY 


PART  ONE 

CHAPTER  I 

THE  PROJECTION  FLAXES 

Section  1.  In  Descriptive  Geometry  the  object  is  chiefly  to  prepare  draw- 
ings as  follows: — 

(a)  Those  which  will  display  or  describe  by  diferent  views  any  object  or 
arrangement  of  lines  or  figures  discussed; 

{h)  Those  which  will,  by  various  analytical  and  constructive  methods  and 
operations,  discover  or  disclose  facts  as  to  shapes,  inclinations,  appear- 
ances, sizes,  etc.;  and 

(c)  Those  which  will  represent  planes  and  how  they  may  be  disposed  to  one 
another. 

The  views  mentioned  above  in  (a)  are  projections,  and  are  made  on  what 
are  called  planes  of  projection.  The  same  projection  planes,  two  in  number,  are 
also  made  use  of  in  the  discussion  of  planes  referred  to  in  (c).  lines  being  drawn 
over  the  planes  of  projection  and  made  to  represent  other  planes  in  various  alti- 
tudes with  respect  to  the  projection  planes. 

The  planes  of  projection  are  the  Horizontal  Plane  and  the  Vertical  Plane. 
These  are  considered  as  being  jlxed,  and  the  lines,  planes,  figures  or  objects  arc 
considcM-ed  as  having  a  relation  to  them-near  or  otherwise  as  to  distance,  inclined 
or  otherwise  as  to  attitude. 

The  drawings  made  either  represent  jjoints,  fines,  figures  or  objects  hx  views 
thrown  perpendicularly  on  to  these  planes  of  projection  (the  H.P.  and  the  V  P 
as  they  arc  commonly  called),  or  they  indicate  the  intersection  of  the  planes  of 
projection  by  lines  and  planes. 

^    In  order  that  the  drawings  ma>-  be  descriptive  of  what  is  under  consideration 
It  IS  usually  necessary  to  have  one  drawin-  of  (he  same  thing  on  each  of  the  planes 


2  DESCRIPTR^E   GEOMETRY 

of  projection,  so  that,  whatever  it  is,  it  may  be  studied  from  different  points  of  view, 
or  better  realized  by  the  study  of  these  drawings  on  the  two  planes  of  projection. 

/;/  the  case  of  projections  it  is  advisable  and  convenient  to  place  one  drawing 
representing  the  view  projected  perpendicularly  on  to  the  H.P.  and  another  draw- 
ing representing  the  view  projected  perpendicularly  on  to  the  V.P.  These  draw- 
ings, projected  perpendicularly  from  whatever  is  under  discussion,  will  arrange 
themselves  perpendicularly  opposite  each  other. 

For  convenience  we  may,  while  dealing  with  simple  and  elementary  problems, 
use  one  part  of  the  drawing  paper,  or  blackboard,  upon  which  to  make  the  draw- 
ings or  projections  thrown  on  to  the  H.P.  These  drawings  are  called  the  hori- 
zontal projections,  or,  more  commonly,  the  plans.  The  other  part  of  the  drawing 
paper,  or  blackboard,  may  be  used  for  making  the  drawings  or  projections  thrown 
on  to  the  V.P.,  called  therefore  the  vertical  projections,  or,  more  commonly,  the 
elevations. 

In  order  to  make  use  of  the  drawing  paper  effectively  in  this  way,  a  Hne  is 
drawn  across  it,  and  this  is  to  represent  the  intersection  line  in  which  the  V.P. 
and  the  H.P.  meet.  We  name  this  line  XY,  and  usually  the  lower  part  of  the 
paper  then  represents  the  near  part  of  the  H.P.,  while  the  upper  part  represents 
the  upper  part  of  the  V.P.  The  line  XY,  however,  must  not  be  thought  to  limit 
the  extent  of  either  of  the  planes,  for  it  may  sometimes  be  necessary  to  display 
something  which  would  not  project  on  to  the  near  part  of  the  H.P.,  and  then 
the  upper  part  of  the  paper  would  also  represent  the  H.P.,  that  is,  the  part  of  it 
beyond  the  V.P.'s  position.  Similarly,  the  lower  part  of  the  paper  may  some- 
times be  required  for  projections  on  the  V.P.  from  things  placed  below  the  level 
of  the  H.P. 

It  must  continually  be  remembered  that  the  part  of  the  drawing  paper  upon 
which  vertical  projections  or  elevations  are  made  is  to  be  considered  actually 
vertical,  and  that  the  part  of  the  paper  used  for  the  horizontal  projections  or 
plans  must  always  be  considered  horizontal.  Sometimes  it  may  be  well  to  fold 
the  drawing  paper  along  the  XY  line  and  then  hold  the  V.P.  part  vertical,  while 
the  H.P.  part  remains  horizontal.     Try  this. 

The  student  must  learn  as  soon  as  possible  to  picture  to  himself,  or  arrange, 
simple  things  in  different  positions  and  attitudes  in  respect  to  the  two  projection 
planes  arranged  as  above  decided  upon,  one  vertical  and  the  other  horizontal, 
and  imagine  what  should  be  the  views  projected  to  these  planes  and  how  the  views 
or  projections  will  change  with  a  change  of  attitude. 

When  it  is  realized  that  this  process  of  projection  is  perpendicular  to  the  pro- 
jection planes  and  not  like  that  of  the  camera,  then  it  will  be  seen  that  the  size 
of  the  projection,  plan  or  elevation,  as  the  case  may  be,  will  not  depend  on  distance. 

The  hand,  resting  on  its  edge  on  the  table  top,  and  arranged  with  the  palm 
vertical,  might  be  used  to  represent  roughly  the  V.P.  of  projection,  and  the  table 
top  to  represent  the  H.P.  of  projection.  Now  hold  things  with  the  other  hand, 
and  consider  what  views  or  appearances  (elevations  and  plans)  you  might  get. 


THE   PROJECTION   PLAXES  3 

projected  perpendicularly  on  to  the  palm  of  the  hand  and  on  to  the  table  top 
respectively.  The  line  across  the  table  top  where  the  vertical  plane  of  the  pahn 
of  the  hand  meets  it,  will  be  the  XY  line  named  above,  and  it  will  be  seen  that 
any  two  drawings  of  an  object  which  is  held  still,  the  one  drawing  on  the  table 
top  and  the  other  on  the  palm  of  the  hand,  are  perpendicularly  opposite  each  other 
across  the  XY  line,  when  the  hand  is  thrown  back  about  the  AT  line  as  a  hinge. 

We  have  seen  that  there  may  be  plans  or  horizontal  projections  and  elevations 
or  vertical  projections  of  points,  lines,  figures  and  objects,  but  it  must  be  pointed 
out  that  there  is  no  such  thing  as  the  plan  or  the  elevation  of  a  plane.  A  plane, 
however,  other  than  the  planes  of  projection,  may  be  represented  by  a  line  or 
lines  drawn  on  the  projection  planes  to  mark  its  intersection  with  them. 


EXERCISE  I 

1.  Distinguish  between  attitude  and  position. 

2.  What  are  plans  and  elevations,  and  how  are  they  obtained? 

3.  What  is  the  XY?  What  is  meant  by  H.P.  and  by  V.P.?  What  is  the  proper  relation 
of  the  planes  of  projection  to  one  another? 

4.  How  can  the  same  sheet  of  paper  serve  for  both  planes  of  projection? 

5.  How  arc  planes,  other  than  the  projection  planes,  represented? 

6.  How  are  a  plan  and  an  elevation  of  an  object  arranged  so  as  to  show  that  they  are  views 
of  the  same  object? 

By  further  investigation  on  the  same  lines  as  suggested  above,  it  will  be  seen 
that  the  distances  of  an  object  held  away  from  the  planes  of  projection  are  the 
same  as  the  distances  of  the  drawings  or  projections  of  it  from  XY.  In  other 
words,  the  plan  at  3"  from  XY  will  mean  that  the  thing  represented  is  3"  from  the 
V.P.,  and  the  elevation  at  2"  from  AT  will  mean  that  the  thing  is  2"  from  the  H.P. 

When  an  object  is  on  the  near  side  of  the  V.P.  it  is  said  to  be  in  front  of  the 
V.P..  and  its  plan,  on  the  paper,  will  be  placed  below  the  AT  line. 


EXERCISE  II 

I.  Let  the  projection  planes,  for  the  purpose  of  this  exercise,  be  represented  by.  say,  a  large 
book,  standing  on  the  table,  for  the  V.P.,  and  by  the  table  top  for  the  H.P..  then  arrange  a  book 
or  other  rectangular  object,  in  such  altitudes,  with  respect  to  the  projection  planes,  as  to  satisfy 
the  following  descriptions: — 

(i)  The  book,  or  tlat    rectangular   bo.\,  placed  horizontally  with  all  its  edges  equally 
inclined  to  the  \'.P. 

(2)  The  same,  so  thai  its  true  form  (rectangle)  will  appear  in  elevation. 

(3)  The  same,  so  that  its  appearance  will  be  a  rhomboid  on  both  planes. 

14)  The  same,  so  that  one  edge  is  in  the  V.P.  at,  say,  45°  to  AT,  and  the  surface  of 
it  is  at,  say,  60°  to  the  V.P. 


4  DESCRIPTIVE   GEOMETRY 

(5)  The  same,  so  that  both  projections  are  lines  only,  supposing  the  book  to  be  very 

thin. 

(6)  The  same,  so  that  its  plan  and  elevation  are  both  rectangles. 

2.  Make  sketch  or  freehand  drawings,  on  a  paper  or  blackboard,  not  carefully  measured 
or  scaled,  as  plans  and  elevations,  to  satisfy  the  descriptions  above,  and  insert  distances  from 
the  planes  of  projection,  wherever  possible. 

3.  How  are  distances  from  the  planes  of  projection  shown,  of  any  points  in  an  object  whose 
plan  and  elevation  are  required  or  given? 


LINES  AND  THEIR  INCLINATIONS 

Section  2.  It  will  novv^  be  seen  that  any  object,  such  as,  for  instance,  a  book, 
may  have  many  different  sets  of  drawings  (a  set  =  plan  and  elevation),  differing 
according  to  its  attitude  or  arrangement  in  relation  to  the  projection  planes,  to 
represent  it,  and  that,  given  any  such  set  of  drawings  for  the  book,  the  book  may 


(i) 

(ii) 

(iii) 

be' 

(iv) 
c           b' 

J  a' 

1  \ 

\    \ 

b'e 

( 

:d:_ 

— 

1      \ 

1 
1 
1 

'c'd' 

i\    \ 

1     !       1     1 
1     1       1     1 

d 

1 

d 

'  /k    ' 

'/\l 

id 

■■'-4:^' 

J 

c 

b      < 

b 

Fic 

I. 

be  held  in  the  proper  way  to  give,  or  provide  for,  that  set  of  drawings  or  pro- 
jections. 

In  order  to  distinguish  plans  from  elevations,  it  is  customary,  for  purposes 
of  discussion,  to  use  small  letters  a,  b,  c,  etc.,  on  the  plans,  and  the  same  small 
letters,  with  a  short  stroke  or  dash  as  a',  h' ,  c' ,  etc.,  for  the  same  points  named 
on  the  elevations.  Capital  letters  A,  B,  C,  etc.,  are  used  for  the  names  of  points 
regardless  of  their  projections. 

Illustrations  of  projections^  or  plans  and  elevations  are  here  given,  which 
must  be  thoughtfully  studied,  while  many  facts  are  noted. 

In  Fig.  I  at  (i)  the  point  A,  is  represented  at  a  and  a'  and  has  a  distance  above 
the  H.P.  equal  to  twice  that  of  its  distance  in  front  of  the  V.P.  At  (ii)  the  square 
BCDE  is  arranged  horizontally  with  edges  BC  and  DE  parallel  to  the  V.P.  The 
distance  of  the  elevation  above  XY  shows  the  distance  of  the  square  above  the 
H.P.     At  (iii)  the  square  still  has  its  edges  BC  and  DE  parallel  to  the  V.P.  and 


LINES   AND   THEIR   INCLINATIONS  5 

its  other  edges  still  horizontal,  i.e.,  parallel  to  the  H.P..  but  the  square  is  inclined 
to  the  H.P.,  and  the  angle  the  elevation  line  makes  with  XY  is  the  angle  the  figure 
makes  with  the  H.P.  The  figure  is,  of  course,  still  perpendicular  to  the  V.P. 
At  (iv)  the  plan,  the  same  shape  and  size  as  thef  plan  at  (iii),  has  been  so  turned 
that  BC  and  DE  are  no  longer  parallel  to  the  V.P.,  but  as  these  edges  are  still 
inclined  the  same  amount  as  before,  to  the  H.P.,  the  difference  of  level  of  their 
ends  has  not  been  altered,  and  so  the  elevation  can  be  derived,  partly,  from  the 
elevation  at  (iii).  Realize  that  the  edges  CD  and  BE  are  still  horizontal,  but 
inclined  to  the  V.P.,  the  angle  of  inclination  to  the  V'.P.  being  that  which  cd  makes 
with  XY. 

N.B. — A  horizontal    line  may  have  any  direction,  for   its   plan,    across  the 
paper. 


Consider  next  the  group  of  drawings  in  Fig.  2.  At  (i)  a  square  is  represented 
horizontal,  at  a  distance  above  the  H.P.  and  a  little  in  front  of  the  V.P.;  one 
diagonal  is  perpendicular  to  the  V.P.  and  is  represented  in  elevation  at  b'd'.  The 
full  length  of  the  diagonal  AC  is  shown  in  both  the  plan  and  the  elevation.  At 
(ii)  the  elevation  of  (i)  has  been  inclined  to  give  an  inclination  of  the  figure  to  the 
H.P.,  the  point  at  b'd'  still  representing  the  horizontal  diagonal  whose  plan  bd 
therefore  shows  true  length;  the  plan  of  the  other  diagonal,  however,  is  now  short 
of  its  true  length,  due  to  its  inclination.  This  diagonal  is  said  to  he  foreshortened, 
that  is,  arranged  so  that  the  view  of  it  does  not  show  its  true  size. 

Now,  if  without  altering  the  size  and  shape  of  the  plan  of  this  square  as  seen  in 
Fig.  2  at  (ii),  it  be  moved  so  that  the  horizontal  diagonal  BD  be  at  an  inclination 
to  the  V.P.  represented  by  ^he  angle  which  bd  makes  with  AT,  it  will  be  seen 
that  the  relative  levels  of  the  various  points  ABCD  will  remain  as  before,  and 
hence  a  new  elevation  for  set  'iii)  may  be  obtained  from  the  plan  as  now  arranged 


6  DESCRIPTIVE   GEOMETRY 

at  (iii)  and  by  making  use  of  the  levels  from  the  elevation  of  (ii).  The  square 
may  now  be  described  as  having  a  diagonal  horizontal  and  at  an  angle  {a°)  to  the 
V.P.,  and  the  figure  as  having  an  inclination  (/3°)  to  the  H.P.  It  will  be  seen 
that  in  order  to  obtain  set  (iii)  it  is  necessary  to  first  make  sets  (i)  and  (ii). 

From  further  inspection  of  Fig.  i  and  Fig.  2  it  will  be  seen  that  a  line,  in  order 
to  show  its  true  length  in  a  projection,  must  be  arranged  parallel  to  the  plane 
upon  which  the  projection  is  thrown;  and,  similarly  with  regard  to  plane  figures, 
i.e.,  for  a  figure  to  show  its  full  size  and  shape  in  a  projection,  it  must  be  arranged 
parallel  to  the  plane  upon  which  the  projection  is  made. 

This  matter  is  still  further  illustrated  and  discussed  in  Fig.  3,  where  it  will 
be  seen  that  the  plans  in  (i)  and  (ii)  and  (iii)  remain  full  length  because  the  line 
is  horizontal  in  each  case,  whereas  the  elevations  vary,  because  the  relation  of  the 
fine  to  the  V.P.  varies. 


(I) 


(III) 


ah 


a'         h'     a' 


b  a  b 


;b  ^^ 


I 


Fig.  3. 


In  (iii)  the  plan  and  elevation  are  both  full  size.  In  (iv)  the  elevation  is 
full  size  because  the  line  is  still  parallel  to  the  V.P.  as  in  (iii),  and  as  shown  by  the 
arrangement  of  its  plan,  now  foreshortened,  however,  because  the  line  is  inchned 
to  the  H.P.  In  (v)  the  plan  is  kept  the  same  length  as  in  (iv)  and  consequently 
the  levels  for  the  ends  of  the  elevation  may  be  taken  from  the  elevation  in  (iv). 
At  (vi)  the  line  is  vertical. 

It  will  readily  be  seen  that  in  Fig.  3  at  (ii)  the  angle  the  line  makes  with  the 
V.P.  is  the  angle  that  ab  makes  with  XY,  and  that  in  (iv)  .the  angle  the  line  makes 
with  the  H.P.  is  the  angle  a'b'  makes  with  XY. 

Realize  that  although  some  of  its  projections  are  inclined  to  XY,  the  fine 
AB  itself  is  not  inclined  to  XY. 

If  now  it  be  required  to  discover  the  angles  th^  line  AB,  in  Fig.  3  as  shown^ 
at  (v),  makes  with  the  projection  planes,  it  is  evident  that  the  angle  it  makes  with 


LIXES    AND   THEIR   INX'LIXATIOXS  7 

the  H.P.  is  not  what  its  elevation  at  (v)  makes  with  XY,  but  what  its  elevation 
at  (iv)  makes  with  AT,  where  the  plan  is  arranged  parallel  to  XY;  and  in  order 
to  find  what  inclination  the  line  AB  has  with  the  V.P.  its  elevation  a'b'  (v)  must 
be  arranged  horizontally  as  at  (ii),  and  then  its  plan,  the  full  length  of  the  line, 
will  show  with  XY  the  angle  required. 

Let  it  be  required  to  fmd  the  true  lengths  of  any  Hnes  AB,  Fig.  4.  and  CD, 
Fig.  5,  and  also  their  inchnations  to  the  planes  of  projection.  Their  plans  and 
elevations  are  shown  at  ab,  a'b',  Fig.  4  and  at  cd,  c'd',  Fig.  5.  The  line  in  each 
case  must  be  arranged  parallel  to  the  V.P.  by  swinging  its  plan  to  become  parallel 
to  XY  as  at  0^2,  Fig.  4,  and  at  tJ..,  Fig.  5,  with  the  result,  when  the  elevation 


Y     X 


Fig.  4. 


opposite  this  new  arrangement  of  the  plan  in  each  case  is  set  up.  that  the  true 
length  d'b'2  and  c'd' 2  respectively,  is  shown  on  the  V.P.,  and  the  incUnation  0^,  of 
the  line,  to  the  H.P.  may  now  be  seen;  also,  the  line  in  each  case  must  be 
arranged  parallel  to  the  H.P.  by  swinging  its  elevation  parallel  to  XY,  and 
so  providing  for  a  plan,  a-ib  and  cod  in  the  cases  being  considered,  that  will 
show  true  length  and  contain  an  angle,  a°,  with  XY,  equal  to  the  angle  the 
hne  makes  with  the  V^P. 

N.B. — The  secondary  positions  for   the  lines  in   Figs.  4  and  5  are  shown  by 
double  lines  to  emphasize  them. 


EXERCISE  III 


I.  Place  the  plan  and  elevation  of  the  line  AB,  making  the  plan  ab  2"  long  and  the  elevation 
3"  long.  Find  the  true  length  of  this  line,  and  also  its  inclinations  to  the  planes  of  projection. 
Mark  the  angles  as  in  Fig.  4  with  arrow-headed  arcs  and  use  a.  to  indicate  the  angle  with  the  V.P., 
and  ii  for  the  angle  with  the  H.P. 

^.  The  plan  cd  is  2"  long.  The  line  CD  is  3"  long.  Draw  an  elevation  for  it,  and  show 
what  angles  the  line -CZ)  makes  with  the  planes  of  projection. 


8  DESCRIPTIVE   GEOMETRY 

3.  The  elevation  of  EF  is  2"  long,  and  is  at  45°  to  XY .     The  line  is  at  30°  to  the  H.P.     Find 
a  plan  for  it,  and  also  determine  its  true  length. 

4,  5  and  6.  Find  the  true  length  of  each  of  the  given  lines  GH,  JK  and  LM,  and  their  indina- 
tions  to  the  planes  of  projection. 


CHAPTER  II 

PLAX  AND  ELEVATION  FORMS  FOR  PLANE  RECTILINEAL  FIGURES 

Section  3.  Suppose  it  be  required  to  lind  the  projections  of  any  plane  figure 
inclined  at  a  given  angle  to  one  of  the  projection  i)lanes  and  having  one  of  its 
edges  in  that  projection  plane  at  a  given  inclination  to  the  XV.  Illustrations 
of  such  a  problem  are  shown  in  Fig.  6  and  Fig.  7. 

In  Fig.  6  a  square  is  chosen  and  is  first  placed,  as  at  (i),  in  the  V.P.  Its  plan 
is  therefore  part  of  the  XY  line.     In  order  that  it  may  swing  out  at  the  required 


Fig.  6. 


Fig.  7. 


angle,  and  yet  leave  one  edge  in  the  V.P.,  it  must  be  arranged  so  that  when  the 
line  representing  it  in  plan  is  moved  to  enclose  the  given  angle  a°  with  A'l'.  one 
edge  of  it,  say  AB,  represented  in  elevation  at  a'b\  is  in  plan  a  point,  as  at  ab, 
and  will  remain  as  a  point  in  AT.  The  shaded  or  "hatched"  rectangle  is  now 
the  elevation  shape  required.  This  projection  shape  is  the  first  thing  to  obtain 
toward  the  solution  of  such  a  problem^as  the  above. 

If  now  this  elevation  shape  be  rearranged  so  that  the  edge  AB  is  at  the  given 
angle  to  the  AT,  as  at  (ii),  then  it  will  be  seen  that  a  rhomboidal  figure  will  be 
obtained  for  plan.  The  distance  of  the  plan  line  cd  in  (ii)  from  A'l'  will  be  the  same 
as  the  distance  it  arrived  at  in  (i\  since  the  elevation  shape  is  the  same  in  both 
cases  and  therefore  the  relation  of  the  figure  to  the  V.P.  is  unchanged. 

9 


10  DESCRIPTIVE   GEOMETRY 

In  Fig.  7  the  case  of  the  regular  pentagon  should  be  studied.  The  pentagon, 
as  l.nally  arranged  at  (ii),  may  be  described  as  being  inclined  at  (3°  to  the  H.P. 
and  having  one  edge  in  the  H.P.  at  a°  to  XY.  The  mode  of  procedure  is  similar 
to  that  for  the  square  in  Fig.  6. 

N.B. — It  is  advisable  that  the  student,  in  order  to  comprehend  clearly  this 
matter  of  projections,  should  frequently  fold  his  drawing  paper  on  the  XY  line 
and  stand  that  part  of  the  paper  with  elevations  on  it,  vertically,  so  that  the  planes 
of  projection  are  rightly  related.  He  will  then  realize  how  the  projections  are 
accounted  for. 

Before  working  Exercise  IV  on  paper,  it  may  be  well  for  the  student  to  cut  a 
square,  and  also  a  rectangle,  out  of  a  piece  of  cardboard,  and  practise  placing 
them  according  to  the  questions.  Realize  that  the  projections  or  views  required 
are  thrown  always  perpendicularly  on  to  the  planes  of  projection. 


EXERCISE  IV 

1.  A  rectangle  2"  by  f"  is  the  plan  for  a  square.  Show  an  elevation  for  it,  (a)  when  an  edge 
of  the  square  is  in  the  H.P.  and  perpendicular  to  XY,  (b)  when  an  edge  in  the  H.P.  is  at  45°  to  XY. 

2.  A  square  of  2"  edge  is  the  plan  for  a  rectangle  having  its  long  edges  3".  Arrange  the 
square  plan  so  that  edges  representing  the  short  edges  of  the  rectangle  are  at  30°  to  the  V.P., 
and  then  obtain  an  elevation. 

3.  Find  the  plan  and  elevation  of  a  square,  of  2"  edge,  inchned  at  60°  to  the  H.P  and  having 
one  edge  horizontal  at  a  distance  of  i"  above  the  H.P.  Place  the  square  so  that  the  horizontal 
edges  are  parallel  to  the  V.P.  and  in  front  of  it. 

4.  Find  the  plan  and  elevation  of  a  2"  square  when  one  diagonal  of  it  is  horizontal  and  at 
45°  to  the  V.P.,  while  the  figure  is  inclined  at  60°  to  the  H.P.  (This  is  the  same  as  saying  that 
one  diagonal  being  horizontal  and  at  45°  to  the  V.P.,  the  other  is  at  60°  to  the  H.P.) 

\  5.  The  plan   of  a   square  is  a  rectangle  2§"  by  f ",  with   the  long  edges   parallel  to  XY. 
Find  an  elevation  for  this  square. 

6.  A  rectangle  2"  by  3"  has  a  long  edge  in  the  V.P.  The  figure  is  inclined  to  the  V.P.  at 
60°.  Show  the  elevation  arranged  so  as  to  represent  the  rectangle  with  its  long  edges  at  30° 
to  the  H.P.,  and  obtain  a  plan  for  it. 


PLAN  AND  ELEVATION  FORMS  FOR  THE  CIRCLE 

Section  4.  A  similar  process  to  that  for  obtaining  projections  of  the  pentagon 
in  Fig.  7  is  used  for  projections  of  the  circle. 

For  the  circle  to  appear  circular  in  projection  it  must  be  placed  parallel  to 
the  projection  plane.  If  it  is  horizontally  arranged,  its  plan  is  a  circle,  and  if  it 
is  placed  parallel  to  the  V.P.  its  elevation  is  circular. 

Hold  some  circular  disc  such  as  a  coin  or  a  piece  of  cardboard  cut  circular, 
and  it  will  be  observed  that,  like  any  other  plane  figure,  it  can  be  represented  by 
a  straight  line  on  either  one  or  the  other,  or  on  both  at  the  same  time,  of  the  planes 


PLAX   AND   ELEVATION   FORMS   FOR   THE  CIRCLE 


11 


of  projection;  also,  that  when  not  {perpendicular  to  a  projection  plane,  nor  p>arallel 
to  it,  the  circle  will  appear  in  projection  as  an  ellipse  with  the  major  axis  equal 
to  the  length  of  the  diameter  of  the  circle.  This  major  axis  represents  the  diameter 
parallel  to  the  projection  plane  upon  which  the  ellipse  appears. 

Let  it  be  required  to  find  the  projections  of  a  circle  inclined  at,  say,  a'  to  the 
H.P.  and  ha\'ing  its  horizontal  diameter  at.  say.  ^'  to  the  V.P. 

In  Fig.  8  at  (i)  a  circle  is  shown,  horizontal,  with  a  line  for  its  elevation.  If 
the  elevation  line  be  rearrange^  at  an  inclination  of  a'  to  A' I'  as  at  (u)  it  wiJI 
represent  the  circle  tilted  or  in^ned  at  a°  to  the  H.P.  The  horizontal  diameter 
AB  which  is  perpendicular  to  the  V.P.  will  remain  full  length  in  plan  at  a2*2-  while 
the  diameter  at  right  angles  to  it.  viz.,  CD,  will  now  be  foreshortened  to  C2d2- 


(U) 


(m) 


Fig.  8. 


A  chord  EF  made  parallel  to  AB  wiU  appear  in  ele\-ation  as  a  pwint,  and  in 
plan  at  €2/2-  Other  horizontal  chords  may  be  made  use  of,  and  eventually  suffi- 
cient points  on  the  circimiference  will  be  thus  secured  in  the  plan  at  u  to  insure 
a  good  shape,  when  drawn  by  passing  a  freehand  cur\-ed  line  through  them,  to 
ser\-e  as  the  edge  of  the  circle  when  the  circle  is  tilted  or  inclined  to  the  H.P. 

Xow  if  this  plan,  at  (ii).  be  turned  so  that  the  horizontal  diameter,  a^^s,  and 
horizontal  chords,  are  no  longer  j)erpendicular  to  the  V.P..  but  at  4"  to  the  V-P. 
ELS  at  <iii),  then  an  elevation,  in  the  shape  of  an  ellipse  instead  oi  2  line,  will  be 
btained.  and  this  b  done  by  deri\-ing  the  levels  for  the  e!  -    us 

points  in  the  cur\-e.  from  the  elevation  line  at  (ii.     Such  ..  .  .-r, 

Tiight  also  be  obtained  by  making  use  of  what  is  called  a  secondar\-  ele\*atioo 
3lane.  represented  as  ha\ing  its  XV  at  any  convenient  distance,  as  at  A'^Fi-     This 


12  DESCRIPTIVE   GEOMETRY 

being  settled,  without  moving  the  plan  around,  as  was  necessary  at  (iii),  the  ele- 
vation (iv)  is  projected  directly  from  the  original  plan  at  (ii),  and  heights  for 
the  various  points  are  transferred  by  measurement  from  the  elevation  at  (ii). 

N.B. — This  secondary  XY  must  make  with  the  plan  of  the  horizontal  diam- 
eter 02^2  at  (ii)  the  same  angle,  /3°,  as  that  contained  between  the  XY  and  the  plan 
asbs  at  (iii).     This  will  mean  that  X2Y2  makes  with  XY  an  angle  90°  — /3°. 

It  will  be  seen  that  the  solution  of  the  problem  requires  a  set  of  drawings 
to  be  made,  either  such  as  (i),  (ii)  and  (iii),  or  a  set  such  as  (i),  (ii)  and  (iv). 


EXERCISE  V 

1.  Arrange  a  regular  pentagon,  of  i|"  edge,  so  that  one  edge  is  in  the  H.P.  parallel  to  XY 
and  the  figure  is  inclined  to  the  H.P.  at  45°.  (Note. — The  angle  at  the  corner  of  a  pentagon 
is  io8°.) 

2.  Incline  a  regular  hexagon,  of  ij"  edge,  at  50°  to  theH.P.,  leaving  one  of  its  edges  in  the 
H.P.  Then  turn  the  plan  so  that  the  edge  in  the  H.P.  is  at  30°  to  XY,  and  show  an  elevation 
for  it  when  so  placed. 

3.  Find  a  plan  and  an  elevation  for  an  equilateral  triangle  inchned  at  60°  to  the  H.P.,  and 
having  one  edge,  2I",  in  the  H.P.  at  60°  to  the  XY  line. 

4.  Find  the  plan  and  elevation  of  a  circle,  2^"  diameter,  when  it  is  inclined  at  50°  to  the 
H.P.  Then  arrange  the  plan  so  that  the  horizontal  diameter  (the  major  axis  of  the  ellipse  found 
for  plan)  is  at  45°  to  the  V.P.,  and  obtain  its  elevation. 

5.  Find  the  plan  of  a  circle  2^"  diameter,  inclined  at  60°  to  the  H.P.,  and  find  an  elevation 
on  a  secondary  vertical  plane,  (with  secondary  XY),  arranging  it  so  that  the  horizontal  diam- 
eter of  the  circle  makes  an  angle  of  60°  with  the  secc    'ary  plane. 

6.  Find  the  plan  and  elevation  of  a  circle,  2^"  diameter,  inclined  at  70°  to  the  V.P.,  and 
having  the  diameter  which  is  parallel  to  the  V.P.  at  45°  to  the  H.P. 


CHAPTER  III 

OBLIQUE  PLANES,  THEIR  TRACES  ANT)  INCITNATIONS 

Section  5.  It  has  been  previously  seen  that  in  order  that  a  line  might  be  set 
up  at  a  given  angle  to  the  H.P.,  or  in  order  to  find  the  incUnation  to  the  H.P.  of 
a  given  Hne,  it  is  necessary  to  arrange  it  parallel  to  the  V.P.,  or  in  it,  and  that  then 
the  projection  in  the  V.P.  makes  with  XY  the  same  angle  as  the  line  makes  with 
H.P.  Also,  that  in  order  that  a  plane  figure  might  be  tilted  or  inclined  to  a  definite 
angle  with  the  H.P.,  it  is  necessary  to  arrange  it  in  a  perpendicular  attitude  to  the 
V.P.,  that  is,  so  that  its  appearance  in  elevation  is  a  line  only,  and  we  have  seen 
that  the  angle  this  Une  makes  with  the  AT  is  the  incHnation  the  plane  figure 
makes  with  the  H.P. 

Similarly,  with  respect  to  the  inclination  of  a  line  or  a  plane  figure  to  the  V.P., 
the  line  must  be  arranged  parallel  to,  or  in,  the  H.P.,  and  the  figure  must  be 
arranged  vertically,  so  that  its  plan  is  a  straight  line  only,  and  the  angle  this 
straight  line  then  makes  with  A'F  is  the  inclination  of  the  given  line  or  plane  figure 
to  the  V.P. 

If,  now,  the  plane  of  any  figi  \  that  is,  the  imaginar}'  plane  of  which  the 
figure  is  a  part,  be  considered,  it  is  clear  that  when  such  a  plane  is  vertical  it  will 
cut  or  intersect  the  V.P.  in  a  vertical  Hne,  and  at  the  same  time  cut  or  intersect 
the  H.P.  in  a  line  that  makes  with  XY  the  angle  that  the  said  plane,  arranged  in 
vertical  attitude,  makes  with  the  V.P.  (The  room  door  opened  will  serve  to 
illustrate  this.) 

Also,  it  will  be  seen  that  the  plane  of  a  figure  arranged  perpendicularly  to 
the  V.P.,  but  incUned  to  the  H.P.,  will  cut  or  intersect  the  H.P.  in  a  line  perpen- 
dicular to  the  XY,  and  will  cut  or  intersect  the  V.P.  in  a  hne  which  makes  with 
XY  the  angle  the  plane  makes  with  the  H.P. 

Two  such  planes  are  represented  in  Fig.  9  at  RST  and  LMX .  The  inter- 
section line  RS  is  made  by  the  vertical  plane  RST  meeting  the  V.P.,  and  the  inter- 
section hne  ST  is  made  by  the  same  plane  meeting  the  H.P.  So  also  the  inter- 
section Unes  LM  and  MX  arc  made  by  the  plane  LMX ,  perpendicular  to  the  \'.P., 
but  inchned  to  the  H.P.,  meeting  the  planes  of  projection. 

These  intersection  lines  made  by  a  plane  meeting  the  planes  of  projection 
are  called  the  traces  of  the  plane,  and  it  is  by  traces  that  the  attitude  or  arrange- 
ment of  any  plane  is  expressed. 

The  traces  found  or  marked  in  the  V.P.  are  called  \'ertical  Traces,  though 
they  are  not  necessarily  vertical  Unes,  as,  for  instance,  LM;  and  the  traces  on  the 

13 


14 


DESCRIPTIVE   GEOMETRY 


H.P.  are  called  Horizontal  Traces,  and  these,  of  course,  are  always  horizontal 
lines.  In  mentioning  these  traces  the  initial  letters  only  are  used  as  V.T.  and 
H.T.,  instead  of  the  whole  words. 

Planes  having  other  attitudes  than  the  two  so  far  discussed,  sometunes  called 
oblique  planes,  are  shown  at  OPQ  and  VWZ. 

The  student  must  reaHze  that  we  are  not  now  deaHng  with  plans  and  eleva- 
tions, but  with  intersections  of  the  projection  planes,  made  by  planes  of  which  there 
can  be  no  plans  and  elevations.  These  imagined  planes  are  indicated  by  their 
traces. 

Reproduce  on  a  paper  with  XY  marked,  these  planes  as  in  Fig.  9.  Fold  the 
paper  along  the  line  XY  so  that  the  V.P.  is  vertical  while  the  H.P.  remains  hori- 
zontal, just  as  the  projection  planes  are  always  supposed  to  be  arranged  in  relation 
to  one  another,  and  it  will  be  realized  that  RS  is  really  at  right  angles  to  ST,  and 
that  LM  is  really  at  right  angles  to  MN.     In  the  other  cases,  also,  the  angle 


Fig.  9. 


which  OP  makes  with  PQ  is  much  less  than  what  it  appears  to  be  when  the  paper 
is  not  so  folded,  and  so  also  with  VW  and  WZ. 

In  Fig.  9,  if  the  Vertical  Traces  RS,  LM,  etc.,  be  produced  downward  below 
XY,  the  productions  will  show  where  the  planes  cut  the  V.P.  below  the  level  of 
the  H.P.,  and  likewise,  if  the  Horizontal  Traces  ST,  MN,  etc.,  be  produced  upward 
beyond  the  XY  line,  it  will  be  seen  where  the  planes  RST,  LMN,  etc.,  cut  the 
H.P.  in  that  part  of  it  beyond  the  XY  where  the  V.P.  passes  through  it.  For 
general  purposes,  however,  it  is  sufficient  to.  represent  the  planes  by  two  Hues 
meeting  in  XY  as  in  Fig.  9. 

The  student  should  now  consider  Fig.  10  and  reaUze  that  whereas  the  plane 
RST  is  perpendicular  to  the  V.P.,  as  seen  by  the  fact  that  its  H.T.  is  perpendicular 
to  XY,  and  shows  its  angle  to  the  H.P.  as  a°,  the  line  AB  is  parallel  to  the  V.P., 
its  plan  being  parallel  to  XY,  and  therefore  its  elevation  shows  its  inclination  to 
the  H.P.  as  i8^ 

The  elevation  a'b' ,  being  arranged  perpendicular  to  the  trace  RS,  the  angle 
j3  is  complement  to  the  angle  a,  that  is  to  say,  if  the  plane  is  inclined  to  the  H.P. 


I 


OBLIQUE   PLANES,   THEIR  TRACES   AND   INCLINATIONS      15 

at  40°,  then  the  line  perpendicular  to  it  is  inclined  at  50°  to  the  H.P.;  and  as 
ST  shows  the  plane  to  be  at  90°  to  the  V.P.,'so,  also,  ab  shows  the  line  to  be  0° 
to  the  V.P.,  and  it  is  conclusively  evident  that  the  line  AB  is  perpendicular  to  the 
plane  RST. 

In  the  case  of  the  plane  LAIN  and  the  line  CD,  similar  conclusions  will  be 
arrived  at,  namely,  that  the  line  CD  is  perpendicular  to  the  plane  LMX,  and  that 
the  inclination  of  the  line  to  the  V.P.  is  complement  to  the  inclination  of  the  plane 
to  the  V.P. 

If  the  student  will  now  take  a  pencil  to  serve  for  a  line  and  arrange  it  per- 
pendicularly to  any  oblique  plane — the  hand  may  be  arranged  to  show  the  attitude 
of  the  oblique  plane — he  will  notice  that  the  projection  on  to  the  H.P.,  i.e.,  the 
plan  of  the  line,  is  at  right  angles  to  the  H.T.  of  the  plane;  and  that  the  projection 
of  the  line  on  to  the  V.P.,  i.e.,  the  elevation  of  the  line,  is  at  the  same  time  per- 


du) 


(IV) 


Fig.  ic. 


pendicular  to  the  V.T.  of  the  plane.  Illustrations  of  perpendicular  lines  to  oblique 
planes  are  shown  at  (iii)  and  (iv)  in  Fig.  10. 

N.B. — The  projections  of  the  line  need  not  cross  the  traces  of  the  plane. 

The  method  of  linding  the  angles  of  inclination  of  any  oblique  plane  whose 
traces  are  given  is  now  obvious,  and  will  depend  upon  the  facts  noted  above, 
namely: — (i)  A  line  perpendicular  to  an  oblique  plane  can  be  shown  by  arranging 
its  elevation  at  right  angles  to  the  V.T.  of  the  plane,  and  its  plan  at  right  angles 
to  the  H.T.  of  the  plane.  (2)  The  inclination  of  the  oblique  plane  to  the  V.P. 
is  complement  to  the  inclination  of  the  line  perpendicular  to  it,  to  the  V.P.,  and 
the  inclination  of  the  oblique  plane  to  the  H.P.  is  complement  to  the  inclination 
of  the  line  perpendicular  to  it.  to  the  H.P. 

That  is  to  say,  to  find  the  inclinations  to  the  planes  of  projection,  made  by 
any  plane  whose  traces  are  given,  represent  a  line  perpendicular  to  the  plane  by 
arranging  its  elevation  perpendicular  to  the  V.T.  and  its  plan  perpendicular  to 


16 


DESCRIPTIVE   GEOMETRY 


the  H.T.,  and  find  the  inclinations  of  this  line  by  the  method  shown  in  Fig.  4, 
then  subtract  these  inclinations  from  90°  in  each  case,  and  the  inclinations  of  the 
plane  are  obtained. 

An  illustration  of  the  method  is  shown  in  Fig.  11,  where,  to  find  the  inclina- 
tions of  the  oblique  plane  RST  a  line  perpendicular  to  it  is  represented  at  a'b',  ab. 


Fig.  it. 


The  angle  a°  is  the  complement  of  the  angle  the  fine  AB  makes  with  the  H.P. 
Therefore  a°  is  the  inclination  of  the  given  plane  RST  to  the  H.P.;  and,  for  similar 
reasons,  the  angle  0°  is  the  inclination  of  the  plane  RST  to  the  V.P 


EXERCISE  VI. 

Find  the  inclinations  to  the  planes  of  projection,  of  the  planes  represented  at  A,  B  and  C, 
mark  with  arrow-headed  arcs  the  angles  required,  and  say  which  angles  they  are. 


N.B. — Take  each  of  the  planes  A,  B  and  C  as  a  separate  problem  and  work  to  a  large  scale. 


TRACES   OF   LINES 


17 


TRACES  OF  LINES 

Sections.  A  line  AB  is  represented  by  its  projections  at  (i)  in  Fig.  12.  It 
will  be  observed  that  the  A  end  of  it  is  a  i)oint  in  the  V.P.  at  a\  and  that  the  B  end 
of  it  is  a  point  in  the  H.P.  at  b.  Any  other  point  in  the  line  may  be  taken,  as  at 
C,  and  it  will  readily  be  seen  that  C  is  at  some  distance  in  front  of  the  V.P.  and 
at  some  distance,  also,  above  the  H.P.  If  the  point  C  could  be  moved  in  the  line 
until  it  got  down  to  the  H.P.  it  would  be  at  a  definite  place  in  the  H.P.,  namely, 
at  b,  and  b'  would  be  the  elevation  of  that  place.  Similarly,  if  the  point  were 
moved  up  the  line  until  it  met  the  V.P.  it  would  arrive  at  a,  a  definite  place  in  the 
V.P.  of  which  a  is  the  plan. 


Fig.   12 


Now  consider  the  points  R  and  5  at  (ii).  Let  R  be  a. place  or  point  on  the 
V.P.  and  5  be  another  place  or  point  on  the  H.P.,  and  let  it  be  required  to  find  the 
actual  distance  from  R  to  5. 

R,  being  a  point  in  the  V.P.,  will  have  its  plan  directly  opposite  to  it  in  XV 
at  r,  and  S,  being  a  point  in  the  H.P.,  will  have  its  elevation  perpendicularly  oppo- 
site to  it  on  the  V.P.,  in  the  AT  line  at  s'.  The  double  lines  in  the  figure  show 
the  plan  and  elevation  for  a  line  joining  R  to  S,  and  its  length  may  now  be  found 
at  Rs'y,  which,  of  course,  represents  the  actual  distance  from  R  to  5. 

Another  illustration  is  shown  at  (iii),  where  point  .1/  is  on  the  V.P.  and  A  is 
on  the  H.P.  Following  the  same  rule,  the  plan  of  point  M  is  in  A'l'  at.;;/,  and  the 
elevation  of  A  is  found  in  A'F  at  n'.  The  line  joining  ;;'  to  .1/  will  give  the  ele- 
vation of  the  line  extending  from  point  M  on  the  V.P.  to  the  point  A  on  the  H.P., 
and  the  line  joining  m  to  N  will  give  the  plan  of  the  same  line.  To  obtain  the 
true  length  of  the  line  whose  projections  arc  so  found,  swing  the  plan  line  mA  into 
AI',  and  the  line  Mn'2,  the  length  obtained,  is  the  distance  from  .1/"  to  A. 


18 


DESCRIPTIVE   GEOMETRY 


EXERCISE  VII 

1.  ]\Iark  a  point  R  on  the  V.P.  2"  above  XY,  and  another  point  5  on  the  H.P.  perpendic- 
larly  opposite  to  it,  at  a  distance  of  i^"  from  XY.     Find  the  real  distance  from  R  to  S. 

2.  Mark  a  point  M  on  the  V.P.  at  a  distance  of  2"  below  XY,  and  another  point,  TV,  on 
the  H.P.  i"  in  front  of  the  V.P.  and  considerably  to  the  right  of  M.  Find  the  true  distance 
between  the  two  points. 

3.  A  point  V  on  the  V.P.  is  2"  above  .YF.  Find  a  point  H  on  the  H.P.,  i|"  in  front  of 
the  V.P.  and  3",  real  distance,  from  the  point  V. 

In  Fig.  13  the  two  projections  of  a  line  are  given  in  each  of  five  different 
arrangements  for  it.    If,  in  each  case,  the  line  be  imagined  to  continue  in  its  direction, 


(i)  .             (ii)                  (iii)                     (iv)                    (v) 

'/       '     '            '    '   "/     1       1   \                  1  \           ';       1    \ 

X      i   i         1      1    !     i    \            \  /!     1    \ 

II           1     1             1      /          \    1            III/                1                1  \    ^  1 

\  1                                                • 

Fig.  13. 


beyond  the  end  of  the  given  line,  in  other  words,  if_  the  given  line  be  produced, 
then,  if  its  direction  brings  it  toward  a  plane  of  projection,  the  point  it  strikes  in 
the  plane  of  projection,  or  the  point  of  intersection  it  makes  with  that  plane,  is 
called  its  trace  on  that  plane. 

The  trace  or  intersection  point  is  called  the  Horizontal  Trace  (H.T.)  or  the 
Vertical  Trace  (V.T.)  according  to  which  plane  it  is  found  on.  Thus  at  (i)  the 
line  AB  has  an  H.T.,  but  not  a  V.T.,  since  the  line  is  parallel  to  the  V.P.  and  there- 
fore cannot  meet  it  in  any  point.  At  (ii)  there  is  a  V.T.  and  no  H.T.  At  (iii) 
the  line  has  both  a  V.T.  and  an  H.T.     So  also  with  (iv)  and  (v). 

The  method  of  arriving  at  the  traces  of  any  line  is  no  doubt  apparent,  namely : — 
Through  points  in  XY  where  the  projections  of  the  line,  or  the  projections  produced, 
meet  it,  draw  perpendiculars;  then  locate  the  H.T.  in  one  of  these  where  the  plan 
line  meets  it,  and  the  V.T.  in  the  other  perpendicular  where  the  elevation  Hne 
meets  it. 

If  the  given  line  has  already  met  a  plane  of  projection,  as  in  (iv),  before  being 


TRACES   OF   LINES 


19 


[)roduced,  then  the  place  in  that  projection  plane  must  be  located  and  marked  as 
the  trace  of  the  line. 

A  special  case  presents  itself  when  the  projections  of  the  line  are  both  per- 
pendicular to  XY,  as  at  ab,  a'b' 

in  Fig.  14.     To  find  the  traces  ^JaT-, 

in  this  case,  first  swing  the  line 
down  to  a  horizontal  attitude 
so  that  a'ob'  represents  it  in 
elevation  and  a2b  in  plan.  Pro- 
duce the  plan  buj  to  c  with  its 
elevation  r'.  It  will  now  be 
seen  that  if  the  line  be  length- 
ened or  produced  sufficiently 
to  make  it  meet  the  V.P.,  its 
elevation  will  have  to  be  added 
to,  so  that  it  will  be  from  b'  to 
c'2,  and  this  latter  point  will  be 
the  V.T.  of  the  given  Hne. 

Similarly,  turn  now  the 
plan  ab  to  the  position  ab-j  and 
obtain  the  elevation  a'b'2  for  it 
when  the  line  is  parallel  to  the 
\'.P.  Produce  the  elevation  to 
d',  and  find  that  the  lengthened 
line  will  need  a  proportionately  lengthened  plan 
which  is  the  H.T.  of  the  given  line. 


Fig.  14. 


Carry   d   by 


arc   to  J2 


KXKRCLSK  VIII 


nA 


^''"  Kind  and  mark  clearly  the  true  lengths,  the  angles  to  the  planes  of  pf^ction,  and  the  traces 
of  thd  lines  whose  projections  are  given  at  .1.  B,  C,  etc.     The  lower  linfes  are,  in  all  the  cases, 

a' 


plans.     Separate  the  dilTcrent  cases  so  that  no  overlapping  ol  the  solutions  will  lake  place,  and 
work  to  a  large  scale.  "  "^ 


CHAPTER  IV 


SHADOWS  OF  LINES 


Section  7.  A  ray  of  light  may  be  represented  by  a  line,  and  for  our  present 
purposes,  such  a  hne,  with  an  arrow-head,  will  serve  to  indicate  the  direction  of 
parallel  rays  of  light  and  will  be  represented  by  projections,  plan  and  elevation,  such 
as  may  be  seen  at  (i)  in  Fig.  15,  the  arrow-line  below  XY  always  being  the  plan. 

Suppose  a  stick  represented  by  a  line  AB  he  held  or  placed  in  various  posi- 
tions and  attitudes  in  relation  to  the  planes  of  projection  as  indicated  by  the  pro- 


CD 


(II)  (III)  ,  (IV; 

6n 


T  I 


(VI) 


(VII) 


!       ab 


abc 


Fig.  15. 


jections  for  it  at  (ii),  (iii),  etc.,  Fig.  15.  A  ray  of  light  parallel  to  the  obHque 
line  marked  with  arrow-heads  at  (i),  striking  the  end  B  will  be  stopped  there, 
and  instead  of  the  ray  reaching  one  of  the  planes  of  projection,  the  shadow  of  B 
will  be  recorded  on  that  plane.  In  other  words,  the  shadow  will  be  the  H.T.  or 
the  V.T.  as  the  case  may  be,  of  the  ray.  See  H  and  V  in  (ii)  and  (iii)  respectively. 
Do  similarly,  with  regard  to  any  other  point,  say  C,  in  the  hne  representing  the 
stick  at  (iv)  and  at  (v) .  Also,  other  points  in  the  Hne  may  be  dealt  with  to  prove  that 
a  straight  Kne  will. have  a  straight  Hne  for  its  shadow  on  any  plane  it  is  cast  upon. 

Notice  that  when  an  end  of  the  line  or  stick  touches  a  plane,  as  in  (vi)  and 
(vii),  the  shadow  will  start  from  that  point,  for,  of  course,  no  ray  can  reach  the 
plane  at  the  point  where  the  line  touches  the  plane,  and  hence  that  point  is  part 
of  the  shadow,  and  the  shadow  of  any  other  point  in  the  Hne  wiU  be  found  in  a  Hne 
continuing  across  the  plane  from  the  touching  point. 

An  examination  of  cases  (v),  Fig.  15,  and  (i).  Fig.  16,  wiH  reveal  the  fact  that 
when  a  line  is  paraHel  to  a  plane  the  shadow  on  that  plane  is  equal  in  length  to  the 
line  itself,  and  paraUel  to  it.      Hence  in  cases  shown  at  (ii)  and  (iii)  in  Fig.  16 

20 


SHADOWS   OF   LINES 


21 


this  fact  can  be  taken  advantage  of,  and  therefore  the  shadow  of  the  point  B  having 
been  found,  from  it  a  parallel  to  the  given  line  is  drawn  for  so  much  of  the  shadow 
as  falls  upon  the  plane  to  which  the  given  line  is  parallel,  and  which  catches  the 
shadow  of  B.  From  the  point  in  XY  where  this  shadow  line  meets  XY,  join  up 
to  the  shadow  of  A  caught  on  the  other  projection  plane. 


In  Fig.  1 6,  (v),  it  will  be  seen  to  be  necessary,  where  the  line  is  an  obHque 
line,  and  part  of  its  shadow  is  caught  by  the  V.P.,  to  first  cast  the  shadow  as  if  it 
fell  completely  on  the  H.P.,  part  of  it  beyond  the  V.P.,  and  then  to  join  up  from 
the  point  where  it  crosses  AT  to  the  shadow  of  its  upper  end  on  the  V.P. 


(II) 


Three  other  cases  are  shown  in  Fig.  17,  where  the  line  is  so  arranged  that  it 
is  neither  parallel  to  one  plane  nor  to  the  other.  In  all  three  cases  the  end  A 
touches  the  V.P.  and  therefore  that  part  of  the  shadow  which  falls  on  the  V.P. 


22 


DESCRIPTIVE   GEOMETRY 


must  start  from  a\  while  in  (i)  and  (ii),  for  the  same  reason,  that  part  of  the  shadow 
faUing  on  the  H.P.  must  start  from  b.  The  only  way  to  obtain  the  direction  of 
the  shadow  as  it  crosses  one  plane  or  the  other,  is  to  find  the  shadow  of  some  inter- 
mediate point,  chosen  at  convenience,  such  as  M  in  (i),  N,  say,  at  one- third  dis- 
tance from  A  in  (ii),  and  O,  say,  at  half  way  between  A  and  B  in  (iii). 

The  shadow  of  the  portion  BM  in  (i)  is  obviously  hm2,  and  because  a  straight 
line  gives  a  straight  line  shadow  on  a  plane,  this  shadow  line  must  be  continued 
until  it  meets  XY ,  from  which  point  in  XY  the  balance  of  the  shadow  can  be 
drawn  to  a  . 

EXERCISE  IX 

I.  Find  the  shadows  on  the  planes  of  projection  for  the  lines  at  .4,  B,  C,  D  and  E.  The 
directions  for  the  rays  are  indicated  by  the  arrow  lines,  the  number  of  degrees  near  the  arrow 
line  showing  at  what  inclination  the  projection  of  the  arrow  line  in  each  case  must  be  made  to 
XY.  In  working  the  exercise,  make  the  elevations  of  the  given  lines  about  25"  long,  and  arrange 
the  cases  so  as  to  avoid  overlapping. 


2.  Make  another  set  of  lines  arranged  in  various  attitudes  somewhat  similar  to  those  here 
given,  and  find  the  shadows  cast  when  the  parallel  rays  of  light  are  directed  towards  the  planes 
of  projection,  their  elevations  making  45°  with  XY  and  their  plans  30^. 


SHADOWS  OF  PLANE  FIGURES 

Section  8.  The  application  of  the  method  of  finding  shadows  of  lines  to  the 
problem  of  finding  shadows  of  plane  figures  is  a  simple  step.  Illustrations  are 
given  in  Fig.  18. 

The  rule  for  shadows  of  fines  that  are  parallel  to  a  projection  plane  upon 
which  they  are  cast  will  save  a  lot  of  work,  as  it  will  only  be  necessary,  if  that 
rule  is  taken  advantage  of,  to  find  the  trace  of  a  ray  through  one  point  (A)  in 
cases  (i),  (ii),  (iv),  (v)  and  (vi),  and  the  H.T.  of  the  ray  through]^  and  the  V.T. 
of  the  ray  through  B  in  case  (iii). 

Study  of  cases  (iv)  and  (v)  will  show  the  obvious  fact  that  a  plane  figure, 
parallel  to  a  plane,  will  cast  its  real  shape,  and  size  also,  for  its  shadow,  on  that 


SHADOWS   OF  PLANE   FIGURES 


23 


plane  to  which  it  is  parallel;  therefore  in  case  (vi)  it  is  necessary  only  to  find  the 
trace  of  the  ray  through  the  centre  A,  and  make  a  circle  with  this  trace  as  centre 
and  radius  the  same  as  that  of  the  circle  casting  the  shadow. 


Fir..   i8. 


Further  illustrations  are  shown  in  Fii^s.  ig,  20  and  21,  dealing  with  plant- 
figures  parallel  to  the  H.P.,  but  casting  only  part  of  the  shadow,  in  each  case, 
on  the  H.P. 

In  Fig.  19  a  square  figure,  parallel  to  the  H.P.,  touches  the  \'.P.  at  one  corner. 
First  find  the  H.T.  of  the  ray  through  A  at  (72,  and 
with  lines  parallel  to  the  edges  of  the  figure,  make 
the  shadow  of  the  square  on  the  H.P.  so  far  as  it 
is  cast  upon  that  plane.  Next  find  the  V.T.  of  the 
ray  through  B  at  b'2  and  finish  the  outhne  of  the 
shadow,  which  must,  of  course,  run  up  to  the  point 
where  the  figure  touches  the  V.P.  In  Fig.  20  a 
circle,  parallel  to  theH.P.,  casts  most  of  its  shadow 
on  to  the  H.P.,  found  by  obtaining  //,  the  H.T.  of 
the  ray  through  the  centre  A,  and  using  it  as  a 
centre  for  a  circle,  or  rather,  the  segment  of  a  circle, 
on  the  H.P.  The  rest  of  the  shadow,  namely,  the 
shadow  of  the  segment  BCD,  will  be  caught  by  the  j.i^   ip, 

V.P.     To  obtain  it,  find  V.T.'s  for  a  number  of  rays 

passing  through  points  on  the  curve  BDC.  Some  such  points  are  marked  at  i,  2, 
3  and  4  in  the  figure.  A  free-hand  curve  is  drawn  on  the  V.P.  through  the  V.T.'s 
of  the  rays  passing  into  these  points. 

In  Fig.  21  the  H.T.  of  the  ray  through  the  centre  .1,  falls  beyond  the  VF, 
providing  for  only  the  smaller  segment  of  a  circle  to  be  cast  as  shadow  on  the 


24 


DESCRIPTIVE  GEOMETRY 


H.P.  Use  this  point  H  as  centre,  as  before,  and  describe  the  circular  part  of  the 
shadow  appearing  on  the  H.P.  in  front  of  the  V.P.,  and  then  proceed  to  find  the 
shadow  cast  on  the  V.P.  as  in  Fig.  20. 


Fig.  21. 


EXERCISE  X 

1.  Find  the  shadows  for  the  rectangle,  the  square  and  the  circles  shown  at  (a),  {h),  (c)  and 
(d).     Keep  them  well  apart  from  each  other. 

2.  Find  the  shadow  of  the  awning  represented  in  plan  and  elevation  at  (e). 


3.  Arrange  other  cases  for  the  square  and  the  circle,  varying  the  direction  of  the  rays  of  light. 
For  example,  let  a  horizontal  circle,  2"  diameter,  touch  the  V.P.  3I"  above  the  XY,  and  let  the 
rays  have  elevations  at  45°  and  plans  at  30°  to  XY.  The  result  will  be  a  shadow  on  the  V.P. 
elliptical  in  shape. 

N.B. — The  student  is  urged  to  verify  these  solutions  by  actual  sunlight,  and  models  held 
in  various  relations  to  the  planes  of  projection  properly  held  at  right  angles  to  one  another. 


CHAPTKR  V 

THE  "COMPOUND  ANGLE  "  FOR  LINES 

Section  9.  Previously  we  have  seen  how  to  find  the  true  length  of  a  given 
line,  and  also  its  inclinations  to  the  projection  planes,  when  the  projections,  plan 
and  elevation,  are  given;  and  it  has  been  realized  that  the  length  made  for 
the  projection  of  a  given  line  will  influence  its  inclination  to  the  projection  plane. 
See  Fig.  4.  So,  conversely,  the  projection  length,  for  a  given  line,  depends  on  its 
real  length  and  its  inclination  to  the  projection  plane.  Illustrate  this  fact  by  the 
use  of  a  stick  or  pencil  to  serve  for  a  line. 


Fig.  2: 


It  must  also  now  be  realized  that  the  conditions  of  its  length  and  of  its  inclina- 
tion to  a  projection  plane,  establish  the  dilTerence  of  distance  of  the  ends  of  the  line 
from  that  projection  plane. 

Thus  in  Fig.  22  at  ^,  a  line  arranged  parallel  to  the  V.P.  will  show  its  true 
length,  say,  2",  and  its  true  inclination  to  the  H.P.,  say  45°,  while  its  plan  length, 
depending  on  these,  is  shown  by  the  arrow-ended  line  P  below  AT,  and  also  the 
difference  of  distance  of  its  ends  from  the  H.P.  may  be  measured  between  the 
levels  of  the  dotted  lines  passing  horizontally  through  the  ends  of  the  elevation, 
and  marked  by  another  arrow-headed  line,  D. 

A  line  of  the  same  length  arranged  at,  say,  30°  to  the  V.P.,  will  show  that  angle 
between  its  plan  and  XY,  and  also  will  show  its  true  length  in  the  plan  if  the  line 
be  placed  as  a  parallel  to  the  H.P.,  as  in  Fig.  22  at  5,  with  similar  results  as  to 
projection  length,  E,  on  the  other  plane,  and  difference  of  distance  of  its  ends,  D. 
from  that  other  plane. 


26 


DESCRIPTIVE   GEOMETRY 


Note  that,  because,  while  the  length  of  the  hne  remains  the  same  at  5  as  at  ^, 
the  angle  is  changed,  therefore  the  length  of  projection  is  changed,  and  also  the 
difference  of  distance  of  its  ends  from  the  projection  plane  is  changed. 

Hence  it  will  be  seen  that  there  are  four  facts,  which  must  be  borne  in  mind, 
about  any  given  hne,  namely: — (a)  Length;  (b)  Inclination — attitude  in  relation 
to  a  projection  plane;  (c)  Projection  length,  and  (d)  Diference  of  distance  of  its 
ends  from  the  projection  plane.  Given  any  two  of  these  facts  the  others  may 
be  found. 

In  further  illustration  of  this,  let  a  Hne,  say  2"  long,  have  its  lower  end  fixed 
in  XY,  and  be  inclined  at,  say,  45°  to  the  H.P.  By  placing  it  in  the  V.P., 
as  in  a  previous  lesson,  the  Hne 
will  show  its  true  length  for 
elevation  and  its  true  inclination 
to  the  H.P.  in  its  projection  on 
the  V.P.  It  win  be  foreshort-  -^ 
ened  in  plan  as  seen  in  Fig.  23 
at  A  to  an  extent  depending  on 
its  true  length,  2",  and  its  in- 
clination, 45°,  to  the  H.P.  In 
other    words,    the    plan    length 

proper  to  this^line  is  found,  viz.,  ab,  and  also  the  level  for  the  free  end  of  the  line 
is  found  to  be  at  a  height  H  above  the  level  of  the  lower  fixed  end.  Again,  if  a 
line,  say  2"  long,  be  arranged  to  lie  in  the  H.P.  at,  say,  30°  to  the  V.P.,  with 
one  end  fixed  in  the  XY  as  at  B  in  the  figure,  then  the  elevation  length,  a'b',  and 
the  proper  distance,  Z>,  from  the  V.P.,  of  the  free  end,  are  determined. 


Fig.  23. 


EXERCISE  XI 


A; 

Find  the  other  projection  for  each  of  the  given  five  lines,  one  projection  of  which  is  given, 
together  with  a  description  as  follows: — 

AB  is  25"  long  and  one  end  of  it  is  in  the  H.P. 
CD  is  inclined  to  the  H.P.  at  30°. 
EF  is  at  60°  to  the  \'.P. 
GH  is  at  45°  to  the  H.P. 
JK  is  at  60°  to  the  V.P. 


THE   "COMPOUND   ANGLE"   FOR   LINES 


27 


By  combining  or  compounding  the  results  we  have  obtained  at  A  and  at  B 
in  Fig.  23  we  shall  arrive  at  a  method  for  the  solution  to  the  problem  of  finding 
the  plan  and  elevation,  properly  placed  in  relation  to  each  other  and  to  XV,  of 
any  given  line  at  given  angles  to  the  planes  of  projection. 

Thus,  as  an  illustration,  let  the  line  be  2"  long,  with  one  end  tlxed  in  XY, 
and  let  the  angle  it  makes  with  the  H.P.  be  45°,  and  that  with  the  V.P.  30^. 

In  Fig.  24  at  (i)  the  height  or  level  for  the  upper  end  of  the  2"  line  inclined 
at  45°  to  the  H.P.  is  marked  by  the  dotted  line  hh,  and  the  elevation  length  for  a 
2"  line  at  30°  to  the  V.P.  is  obtained  and  marked  e'e'.  This  line  is  then  moved 
by  an  arc  into  its  proper  position,  namely,  between  the  levels  proper  for  a  2"  line 
at  45°  to  the  H.P.,  i.e.,  between  XY  and  the  parallel  ////.  Now  as  the  projections 
of  any  line  are  necessarily  perpendicularly  opposite  each  other  across  XY,  it  follows 
that  the  plan  for  this  line  at  A  must  be  opposite  its  elevation,  and  as  it  must  also 


(in 


(III) 


Fio.  24. 


be  limited  to  the  space  between  XY  and  the  parallel  dd  due  to  the  30''  angle,  a 
perpendicular  from  the  free  end,  that  is,  the  upper  end  of  the  elevation,  drawn  to 
this  parallel  dd,  will  decide  its  plan. 

At  (ii)  the  same  result  is  arrived  at  by  obtaining  the  plan  length,  marked  pp, 
and  moving  it  by  an  arc  into  its  position  due  to  the  30^  angle,  and  from  this  plan 
the  elevation,  which  must  be  opposite  to  it,  is  found. 

If,  instead  of  the  arrangement  of  the  line  by  which  its  projections  meet  in  a 
point  in  AT  as  in  Fig.  24  at  (i)  and  (ii),  an  arrangement  of  it  is  desired,  so  that  its 
upper  end  is,  say,  in  the  V.P.  and  its  lower  end  is,  say,  in  the  H.P.,  then  the  solu- 
tion will  be  obtained  most  readily  by  working  it  out  as  at  (iii),  in  the  following 
order: — (a)  plan  length  due  to  45°;  (b)  distance  of  lower  end  from  AT  due  to  30"; 
(c)  plan  put  into  position,  and  (d)  elevation  found. 

This  problem,  discussed  in  Fig.  24,  is  sometimes  spoken  of  as  the  problem  of 
the  Compound  Angle. 

In  Fig.  25  are  shown  two  common  cases  of  error  made  by  beginners  in  their 
attempts  to  solve  this  problem  without  properly  understanding  what  they  are 


28  DESCRIPTIVE   GEOMETRY 

doing.  The  student  is  urged  to  make  frequent  use  of  some  commonplace  model 
of  the  planes  of  projection,  arranged  at  right  angles  to  one  another,  and  to  mark 
upon  them  projections  of  lines  and  to  study  these.  Then  open  out  the  90°  angle 
between  the  planes,  and  so  represent  the  flat  paper  upon  which  the  projections 
are  to  be  made.     Eventually  the  student  will  be  readily  able  to  "read"  sets  of 


Fig.  25. 


projections  and  also  to  imagine  or  picture  them,  for  cases  described,  without  having 
to  VianHlp  a  moHel. 


to  handle  a  model 


EXERCISE  XII 

Find  projections  for  the  following  lines: — 

(i)  AB,  2\"  long,  inclined  at  45°  to  the  H.P.  and  at  30°  to  the  V.P. 

(2)  CD,  2\"  long,  inclined  at  50°  to  the  H.P.  and  at  20°  to  the  V.P. 

In  both  the  above  cases  show  two  different  ways  of  dealing  with  the  construction. 

(3)  EF,  2\"  long,  at  40°  to  the  H.P.  and  at  50°  to  the  V.P. 

(4)  Find  projections  for  a  2\"  line,  at  40°  to  the  H.P.  and  at  30°  to  the  V.P.,  by  a  method 

providing  for  one  end  of  the  Hne  to  be  in  the  H.P.  and  the  other  in  the  V.P. 

(5)  Propose  other  cases,  varying  the  position  of,  say,  the  lower  end  of  the  line. 

N.B. — When  proposing  new  cases  for  oblique  lines,  the  student  should  realize  that  the  sum 
of  the  two  angles  for  the  same  line  must  not  exceed  90°,  and  that  in  the  case  of  the  sum  of  the 
two  angles  being  the  limit  of  90°,  the  plan  and  elevation  will  both  appear  as  perpendicular  to  XY 


It  will  now  be  recognized  that  upon  this  problem  depends  a  method  of  finding 
or  arranging  traces  for  any  oblique  plane  whose  angles  with  the  projection  planes 
are  given.  For  instance,  if  it  be  required  to  find  the  traces  for  a  plane  which  is 
inclined  at  50°  to  the  H.P.  and  at  70°  to  the  V.P.,  since  the  traces  required  are 
perpendicular  to  the  projections  of  a  line  that  is  at  right  angles  to  the  plane,  it 
will  be  seen  that  a  Hne  at  40°  to  the  H.P.  and  at  20°  to  the  V.P.  will  be  a  perpen- 
dicular to  this  plane;  therefore,  find  the  projections  of  this  40°,  20°  line,  and  then 
make  traces  for  the  required  50°,  70°  plane,  perpendicular  to  these  projections. 


PROJECTIONS   OF   PLANE   FIGURES 


29 


EXERCISE  Xlir 

Find  the  traces  for  the  following  oblique  planes: — 

(a)  60°  to  the  H.P.  and  50^  to  the  V.P. 

(b)  45°  to  the  H.P.  and  65°  to  the  \'.P. 

(c)  30°  to  the  H.P.  and  60°  to  the  V.P. 

N.B. — Here  it  should  be  noticed,  in  proposing  additional  exercises,  that  the  sum  of  the  two 
angles  for  an  oblique  plane  cannot  be  less  than  qo°,  and  in  the  case  of  the  sum  being  equal  to  90" 
the  traces  for  the  plane  will  run  parallel  to  the  XV  line. 


PROJECTIOXS  OF  PL.VNE  FIGURE.S,  IXVOLVIXO  TflE  COMPOFXI)  AXGLE 

Section  10.  Illustration  may  now  be  given  of  the  application  of  the  com- 
pound angle  problem  to  cases  of  plane  figures  in  which  it  is  involved.  And  first 
let  it  be  realized,  experimentally,  that  when  two  lines  meet  at  right  angles,  if  one 
of  them  is  in  a  plane  of  projection  or  parallel  to  it,  the  projection  on  that  plane, 


Fig.  26, 


Fig.  27. 


of  the  second  line,  will  be  perpendicular  to  the  projection  of  the  first  Une.  Con- 
versel}-,  if  one  of  them  has  to  be  inclined  to  the  planes  of  projection,  the  arrange- 
ment of  the  other  must  follow  it  at  right  angles  to  the  projection  of  the  first,  on 
the  plane  to  which  the  second  one  is  parallel.  Consider,  for  example,  that  if  one 
diagonal  of  a  square  be  horizontal  and  the  other  be  at  45°  to  the  H.P.  and  30° 
to  the  V.P.,  then  it  is  impossible  to  say  that  the  horizontal  one  makes  any  par- 
ticular angle  with  the  V.P.,  for  its  plan  simply  follows  at  right  angles  the  plan  of 
the  inclined  diagonal.  It  is  only  possible  therefore  to  say  of  it  that  it  is  horizontal. 
Study  the  cases  shown  in  Figs.  26  and  27.  The  plan  and  elevation  in  Fig.  26 
have  been  found  for  a  square  having  one  edge,  AB,  in  the  H.P.,  and  an  adjacent 
edge,  BC,  inclined  at  50°  to  the  H.P.  and  at  30°  to  the  V.P.;  and  in  Fig.  27  the 
plan  and  elevation  have  been  found  for  a  square  when  one  diagonal.  AR,  is  hori- 


30  DESCRIPTIVE   GEOMETRY 

zontal  and  the  other  diagonal  is  at  a  compound  angle— 50°  to  the  H.P.  and  30°  to 
the  V.P. 

The  first  step  is  to  find  the  plan  shaj  £,  as  at  (i),  quite  irrespective  of  the  com- 
pounding of  angles  for  the  incHned  edge  or  diagonal.  The  next  step  is  to  find  the 
compound  an^le  for  the  obhque  line— the  edge  BC  in  Fig.  26  and  the  diagonal 
CD  in  Fig.  27.  For  this,  any  length  of  Hne  may  be  taken  to  serve  the  purpose, 
because  all  that  is  required  is  to  obtain  the  direction  of  the  plan,  marked  by  the 
arrow  line  in  each  case.  Now,  either  fit  the  plan  shape'  to  this  arrow  line,  as  at 
J2C2,  Fig.  26,  or  make  C2d2  parallel  to  it  as  in  Fig.  27.  The  elevation,  in  each  case, 
is  the  last  thing  to  find,  and  It  may  be  derived  from  the  newly  arranged  plan,  and 
the  first  elevation  which  provides  levels. 

Care  must  be  taken,  that,  in  working  out  the  compound  angle,  the  same 
length  of  line  must  be  used  (any  length  will  do)  to  be  placed  at  30°  to  XY,  as  that 
which  has  been  placed  at  45°  to  XF. 


EXERCISE  XIV 

1.  Find  the  plan  and  elevation  of  a  rectangle,  2"  by  3",  when  a  short  edge  is  in  the  H.P. 
and  the  long  edges  are  inclined  at  40°  to  the  H.P.  and  at  25°  to  the  V.P. 

2.  Find  the  plan  and  elevation  of  a  square  2"  edge,  when  one  diagonal  is  horizontal  and  the 
other  is  at  50°  to  the  H.P.  and  at  25°  to  the  V.P. 

3.  Find  the  plan  and  elevation  for  a  circle  2"  diameter,  when  the  diameter. perpendicular 
to  the  horizontal  diameter  (there  is  always  a  horizontal  one)  is  at  55°  to  the  H.P.  and  at  20°  to 
the  V.P. 

4.  Propose,  and  work  out,  other  problems  of  plane  figures  to  involve  the  compound  angle, 
and  realize  that  questions  like  Nos.  5  and  6,  which  follow,  do  not  involve  the  problem  of  finding 
the  compound  angle. 

5.  Find  the  plan  and  elevationwaf  a  rectangle  having  its  long  edges  horizontal  and  inclined 
at  30°  to  the  V.P.,  and.its  short  edges  at  50°  to  the  H.P. 

6.  Find  the  plan  and  elevation  of  a  square  when  c  ne  diagonal  is  horizontal  and  at  50°  to  the 
V.P.  and  the  other  diagonal  is  at  25°  to  the  V.P. 


(  HAPTKR   \l 

PROJECTIOX  OF  SIMI'LK  SOLIDS 

SectiOx\  11.  The  student  will  now  be  able  to  deal  with  easy  problems 
involving  the  use  of  simple  solids,  and  as  the  solids  used  will  be  familiar  tj-pes,  the 
attention  can  be  chiefly  given  to  the  descriptions  with  regard  to  position  or  place, 
and  with  regard  to  attitude  while  in  any  particular  position.  The  student  is 
advised  to  make  use  of  small  cardboard  boxes  or  models  of  wood,  and  continu- 


FlG.    28. 


Fig.  29. 


Fig.  30. 


ally  to  remember  that  we  are  to  represent  these  solids  by  projecting  their  appear- 
ances perpendicularly  on  to  the  two  planes  of  projection.  Therefore  he  should 
study  the  possible  attitudes  or  arrangements  of  the  solids  in  relation  to  a  vertica- 
plane  and  a  horizontal  plane  held  or  placed  conveniently  for  experimenting  purl 
poses. 

In  the  illustrations  Figs.  28,  29  and  30.  the  order  of  procedure  may  be  studied 
for  finding  plans  and  elevations  of  such  solids  referred  to  above. 

In  Fig.  28  is  represented  a  cube  having  one  of  its  faces  in  the  H.P.  and  another 
face  at  30°  to  the  V.P.  The  dotted  line  in  the  elevation  represents  an  edge 
not  seen,  since  the  student  is  supposed  to  look  toward  the  projection  plane  with 
the  solid  between  him  and  the  projection  plane.  This  applies  to  both  planes, 
hence  the  dotted  lines  appearing  both  in  plans  and  in  elevations  in  Figs.  29  and  30. 
Note  that  the  same  edges  are  not  represented  by  dotted  lines  in  both  projections. 

31 


32 


DESCRIPTIVE   GEOMETRY 


In  Fig.  29  the  cube  has  one  edge  in  the  H.P.  at  30°  to  XY,  and  a  face  adjacent, 
that  is,  to  which  that  edge  belongs,  is  at  30°  to  the  H.P.  There  must  evidently 
be  a  preliminary  set  of  projections  at  (i)  before  the  plan  and  elevation  required 
can  be  found  at  (ii).  This  also  applies  to  Fig.  30,  where  the  cube  has  one  edge  of  a 
face  in  the  H.P.  and  another  edge  of  the  same  face  is  at  60°  to  the  H.P.  and  at 
25°  to  the  V.P.  The  preUminary  plan  shape  and  elevation  for  levels  are  shown 
at  (i),  and  the  compound  angle  problem  for  the  edge  BC  is  worked  out  as  in  Fig.  26, 
the  arrow  line  showing  the  direction  for  &2C2  in  the  required  plan. 

In  the  illustration  Fig.  31  a  regular  hexagonal  prism  is  arranged  so  that  a 
short  edge  is  in  the  H.P.  and  the  long  edges  are  at  a  compound  angle,  viz.,  45° 
to  the  H.P.  and  20°  to  the  V.P.     The  upper  end  of  the  solid  is  turned  away  from 


Fig.  31, 


the  V.P.,  and  is  therefore  fully  visible  as  a  foreshortened  hexagon,  while  dotted 
lines  have  to  be  used  for  part  of  the  outline  of  the  lower  end.  In  working  the 
compound  angle  problem  it  is  only  necessary  to  find  the  plan  direction,  marked 
with  the  arrow  line  in  the  figure,  as  the  elevation  will  naturally  assume  its  proper 
direction. 

In  Fig.  32  a  right  cylinder  is  arranged  so  that  its  axis  is  at  50°  to  the  H.P. 
and  at  15°  to  the  V.P.  This  means  that  the  circular  end  will  be  at  40°  to  the 
H.P.  Its  plan  should  be  found  by  the  method  previously  explained.  The  com- 
pound angle  for  any  length  of  line  RR  is  found  in  such  a  way  (see  Fig.  24,  (iii)) 
that  when  the  arrow  line,  which  indicates  the  direction  for  the  plan  of  the  axis, 
is  obtained,  and  the  plan  of  the  axis  of  the  cylinder  is  made  parallel  to  it,  the  lower 
end  of  the  cylinder  is  turned  away  from  the  V.P.  and  appears  in  the  vertical  pro- 
jection in  full  view,  therefore  is  clear-Hned  all  around.  The  upper  end  will  be 
partly  outlined  by  a  dotted  line.     The  elevation  of  the  axis  should  be  drawn, 


PROJECTION   OF   SIMPLE  SOLIDS 


33 


and,  of  course,  will  not  be  at  50'^  to  XY.  When  the  elevation  ellipses  have  been 
correctly  drawn,  the  width  of  the  elevation  of  the  cylinder,  that  is,  at  right  angles 
to  the  axis,  will  be  the  same  as  that  of  the  plan. 


Fig.  3j. 

In  Fig.  33  a  regular  pentagonal  pyramid  is  made  to  rest  with  one  of  its  tri- 
angular faces  in  the  H.P.  and  the  axis  of  the  solid — the  line  from  its  apex  to  the 
centre  of  its  base — is  at  30°  to  the  V.P.      This  means  that  the  compound   angle 


Fig. 


must  be  found  for  the  axis,  since  there  are  two  angles  to  be  taken  into  consideration, 
namely,  the  angle  a°  which  it  makes  with  the  H.P..  and  the  30^  angle  it  makes  with 
the  \'.P.  In  the  fmal  position,  therefore,  the  plan  oi  the  axis  is  made  parallel  to 
the  arrow  line  obtained  by  compounding  these  angles. 


34 


DESCRIPTIVE    GEOMETRY 


In  Fig.  34  there  is  no  compound  angle  problem  required  to  be  worked  out. 
A  right  circular  cone  is  represented  with  its  axis  at  45°  to  the  H.P.,  and  the  hori- 
zontal diameter  of  its  base  is  at  50°  to  the  V.P.  Notice  that  in  drawing  the  lines 
for  the  sides  of  the  incHned  cone,  where  an  ellipse  is  part  of  the  drawing,  these  side 
lines  are  tangent  to  the  curve  of  the  ellipse,  and  do  not  run  into  the  ends  of  the 
major  axis  of  the  ellipse.     It  will  be  advisable  to  draw  the  line  ee  the  full  length 


Fig.  34. 


of  the  diameter  of  the  base,  at  right  angles  to  the  elevation  obtained  of  the  axis 
of  the  cone.  In  finding  the  plans  and  elevations  of  circles  inclined,  not  less  than 
eight  points  in  the  curve  should  be  found. 

EXERCISE  XV 


1.  Find  the  plan  and  elevation  for  a  cube,  2"  edge,  which  has  one  edge  in  the  H.P.  at  60° 
to  XY,  and  a  face  adjacent  to  that  edge  at  30°  to  the  H.P. 

2.  A  pentagonal  prism  has  one  short  edge,  ij"  long,  in  the  H.P.  The  long  edges,  3"  long, 
are  at  45°  to  the  H.P.  and  at  25°  to  the  V.P.     Find  the  plan  and  elevation. 

3.  A  hexagonal  pyramid,  base  edge  i",  axis  3",  has  one  triangular  face  in  the  H.P.,  and 
the  horizontal  diagonal  of  the  base  is  at  45°  to  the  V.P.     Find  the  plan  and  elevation. 

4.  A  right  circular  cone,  diameter  of  base  2",  axis  3",  lies  on  its  side  on  the  H.P.,  with  its 
apex  therefore  in  the  H.P.     Find  the  projections  for  it  when  its  axis  is  at  30°  to  the  V.P. 

5.  Propose  and  work  out  other  problems,  for  example,  on  the  square  prism,  the  cylinder, 
the  pentagonal  pyramid  and  the  cube,  after  the  manner  of  those  already  given,  some  to  involve 
the  problem  of  the  compound  angle  and  some  not. 

N.B. — This  suggestion  that  the  student  himself  shall  be  required  to  prepare  in  writing  the 
exact  description  and  data  for  additional  exercises,  and  then  work  them  out,  is  very  important, 
for  it  is  a  good  test  of  the  student's  grasp  of  the  subject  so  far  as  the  ground  has  been  covered, 
and  his  powers  of  concentration. 


SHADOWS   OF   SOLIDS   AND   OF   GROUPS   OF   SOLIDS 


35 


SHADOWS  OF  SOLIDS  AND  OF  GROUPS  OF  SOLIDS 

Section  12.  The  more  difficult  cases  of  shadow  problems,  involving  the  use 
of  solids,  may  now  be  undertaken.  Let  the  rays  of  light  be  parallel  as  previously, 
and  directed  toward  the  planes  of  projection,  the  direction  to  be  shown  b)-  the 
plan  and  elevation  of  an  arrow  line. 

Before  commencing  to  work  out  the  solutions  for  such  shadow  problems 
as  now  to  be  dealt  with,  the  student  must  endeavor  to  picture  or  imagine  the  solid 
or  group  represented  by  the  plan  and  elevation,  or  described  verbally,  with  the 
light  directed  toward  it,  and  then  try  to  decide  what  edges  will  cause  or  are  likely 
to  cause  the  outline  or  contour  of  the  shadow  which  will  be  cast. 


Fig.  35- 


The  student  will  not  be  able  to  make  very  satisfactory  progress  until  he  can 
readily  conceive  mental  pictures  of  what  he  has  to  deal  with,  realizing  the  relation 
of  what  he  is  picturing  to  the  vertical  and  horizontal  planes  of  projection,  and 
judging  what  the  projections  and  other  results  might  be  like.  Therefore  it  is 
important  that  the  student  should  take  advantage  of  every  opportunity  which 
presents  itself  to  train  himself  in  these  matters. 

In  Fig.  35  it  will  be  seen  that  a  cube  is  represented,  and  by  studying  it  as 
suggested  above,  it  will  be  concluded  that  the  light  will  fall  on  three  of  its  faces, 
and  that  two  vertical  edges,  those  at  .1  and  at  C,  and  four  horizontal  edges,  the 
upper  ones  AB  and  BC,  and  the  lower  ones  AD  and  DC,  are  the  edges  which  will 
cast  shadows  to  give  the  shadow  shape.  Deal  with  them  in  the  order  suggested 
by  the  numbers  i  to  6  in  the  figure.  Notice  that  3  and  part  of  4  will  be  parallel 
to  DC  and  DA  respective!} . 

In  Fig.  36  it  will  be  seen  necessary  to  cast  the  shadow  of  the  apex  of  the  pyra- 
mid on  to  the  H.P.  beyond  the  A'F  line  first,  in  order  to  draw  from  that  point, 


36 


DESCRIPTIVE   GEOMETRY 


on  the  H.P.  the  Unes  forming  the  contour  for  that  part  of  the  shadow  falhng  upon 
the  H.P.  in  front  of  the  V.P.  (see  Fig.  i6,  (v)).  These  contour  Hnes  are  then  con- 
tinued up  the  V.P.  from  where  they  break  at  XY,  to  the  actual  shadow,  on  the 
V.P.,  of  the  apex.     Notice  that  as  two  triangular  faces  of  the  pyramid,  visible 


Fig.  36. 


Fig.  37. 


in  plan,  have  no  light  falling  on  them,  they  are  shaded.  They  are  said  to  be  in 
shade,  as  distinguished  from  the  cast  shadow  just  found.  In  the  case  of  the  cone 
at  Fig.  37,  the  part  in  shade  is  determined  by  the  radii  that  are  perpendicular  to 
the  tangent  hnes  serving  as  contour  or  boundary  Unes  for  the  shadow  on  the  H.P. 


In  Fig.  38  it  must  first  be  realized  that  the  outhne  of  the  shadow  of  the  circular 
block  there  represented  will  be  caused  by  the  vertical  hnes  on  its  sides  found  in 
plan  at  A  and  at  B,  the  upper  rim  ACB,  and  the  lower  run  ADB.  The  shadows 
of  the  vertical  lines  at  A  and  B  should  first  be  found,  then,  by  taking  points  on 


SHADOWS    OF   SOLIDS   AND   OF   GROUPS   OF   SOLIDS         37 

the  rim  lines  mentioned,  the  curves  can  be  obtained.  The  part  of  the  lower  rim 
casting  its  shadow  on  the  H.P.  will  give  a  circular  shadow  with  its  centre  at  e. 
Only  this,  and  the  straight-Hne  parts  of  the  shadow  have  been  found  in  Fig.  38; 
the  remainder  has  been  purposely  left  unfinished  in  order  to  emphasize  the  impor- 
tance of  first  obtaining  correctly  the  parts  that  are  here  found.  In  each  of  the 
cases,  Figs.  35  to  38,  the  part  of  the  surface  in  shade  and  visible  in  the  elevation 
is  shaded  with  hatching  to  indicate  that  fact. 

Fig.  39  represents  a  flat  block  or  slab  resting  on  a  vertical  cylinder.  First 
find  the  complete  shadow  of  the  group  as  it  is  cast  on  the  planes  of  projection, 
and,  in  doing  this,  after  drawing  the  shadow  outlines  on  the  H.P.  and  starting 


Fig.  39. 


Fig.  40. 


them  vertically  up  the  V.P.,  find  the  outline  of  that  part  of  the  shadow  which 
falls  on  the  V.P.  The  best  order  in  which  the  parts  of  the  outhne  are  obtained 
is  indicated  by  the  numbers.  Notice  that  5  will  be  made  parallel  to  4.  and  6  made 
parallel  to  3,  for  reasons  that  are  obvious. 

After  this  shadow  on  the  planes  of  projection  is  found,  then  find  the  shadow 
cast  on  the  near  or  front  half  of  the  cylinder,  by  a  portion  of  the  upper  solid. 
The  group  is  reproduced  in  Fig.  40  in  order  to  avoid  confusion  of  lines,  and  to 
show  only  the  shade  and  shadow  seen  on  the  near  part  of  the  cylindrical  solid. 
The  line  AB  should  first  be  drawn,  perpendicular  to  the  plan  direction  of  the  rays. 
Then  from  B  a  vertical  line  on  the  near  or  front  surface  of  the  cvlinder  in  elevation 


38 


DESCRIPTIVE   GEOMETRY 


divides  the  light  from  the  shade.  Points  in  the  order  i  to  8  should  then  be  decided 
on,  and  shadows  of  these  points  cast  on  the  surface  of  the  cyhnder.  Freehand 
curves  will  then  mark  the  outline  of  the  shadow  cast.  Notice  that  point  2  is  the 
one  that  casts  a  shadow  on  the  outside  generator  whose  plan  is  C. 

In  Fig.  41  a  group  consisting  of  a  pyramid  standing  on  a  square  block  is  repre- 
sented. The  shadow  caught  by  the  top  surface  of  the  block  will  be  obtained  by 
supposing  the  surface  of  it  extended  sufficiently  to  hold  the  whole  shadow.  This 
will  give  lines  i  and  2.     The  points  a  and  h  must  now  follow  the  shadows  of  the 


Fig. 


edges  they  are  on,  to  az  and  ^2.  From  ^2  the  shadow  of  the  slant  edge  of  the  pyra- 
mid continues  on  the  H.P.,  parallel  to  2  on  the  upper  surface  of  the  block,  until 
it  reaches  the  XY ,  when  it  runs  up  the  V.P.  to  the  shadow  of  the  apex,  on  the  V.P., 
and  from  ao  a  line  on  the  V.P.  to  the  shadow  of  the  apex,  completes  the  contour 
or  outline  of  the  shadow  cast. 


EXERCISE  XVI 


1.  Work  out  carefully  all  the  cases  dealt  with  in  Figs.  35  to  41,  making  much  larger  drawings. 

2.  Find  shadows  on  the  H.P.,  the  V.P.  and  on  the  under  solid,  for  a  group  consisting  of  a 
circular  slab  2"  diameter,  f"  thick,  resting  centrally  on  a  cube,  if"  edge.  The  cube  rests  on 
the  H.P.  with  its  vertical  faces  equally  inclined  to  the  V.P.,  and  one  vertical  edge  \"  from  the 
V.P.     The  rays  of  light  have  plans  at  60°  to  XY,  and  elevations  at  45°  to  XY. 

3.  A  square  block  2"  edge,  and  1"  thick,  rests  with  a  square  face  centrally  on  a  circular 
cylinder,  if"  diameter,  2\"  high,  standing  on  the  H.P.    The  horizontal  edges  of  the  square 


SHADOWS   OF   SOLIDS   AND   OF    GROUPS   OF   SOLIDS 


39 


block  are  equally  inclined  to  the  \-.P.     A  short  edge  of  the  upper  sol.d  is  m  the  \  .P.     The  real 
inclinations  of  the  rays  are  50°  to  the  H.i>.  and  30°  to  the  X.V.     Find  the  shade  and  shadows 

4  \  right  circular  cone,  2"  high,  2"  diameter  of  base,  stands  vertically  on  the  centre  of  a 
square  block,  i"  thick,  sides  3"-  The  block  has  one  rectangular  face  parallel  to  the  \  .1  ^and 
i''  from  it.    The  rays  of  light  have  plans  at  45°  to  XY  and  elevations  at  30    to  A  Y.     I-md  the 

shade  and  shadows.  r  wu  , 

5.  Prepare  other  exercises,  using  your   best  judgment,  and   work    them   out    for   further 

practice. 


PART  TWO 


CHAPTER  VII 

RABATTEMENT  OF    PLANES,    AND    PROJECTIONS    OF    FIGURES 
INVOLVING    ITS    USE 

Section  13,  The  process  known  as  rabattement  is  that  by  which  a  plane  is 
made  to  swing  on  one  of  its  traces,  as  on  a  hinge,  until  whatever  is  represented  as 
being  in  the  plane,  whether  it  be  point,  line  or  figure,  is  turned  into  one  of  the 
planes  of  projection,  and  its  true  form  is  made  evident,  also  its  relation  to  the  trace  of 


Fig.  42. 


the  plane.  In  Fig.  42  the  plane  RST,  arranged  perpendicularly  to  the  V.P.,  has 
in  it  a  point  P.  If  the  plane  is  made  to  swing  about  its  trace  5r  as  a  hinge  line 
until  it  is  turned  over  into  the  H.P.,  either  on  the  one  side  or  on  the  other,  it  carries 
the  point  with  it,  and  the  point's  rabattement  is  said  to  be  at  P2  or  at  P3.  If  D  be 
a  point  perpendicularly  opposite  P,  and  in  the  H.T.  or  hinge  line,  then  DP  may 
be  considered  as  the  radius  line  for  an  arc  in  which  the  point  P  moves.  This 
radius  length,  as  shown  in  the  figure  at  (ii),  is  the  hypotenuse  of  a  right-angled 
triangle  having  for  its  base  the  perpendicular  distance  from  the  plan  p  to  the 

40 


RABATTEMENT   OF  PLANES 


41 


trace  at  d,  and  for  its  height  the  distance  from  A'l'  to  p' .  The  right  angle  at  c  in 
the  figure  should  be  specially  noted. 

In  Fig.  43  an  oblique  plane  is  represented  at  RST  and  a  point  P  in  it  with 
plan  p  and  elevation  p' .  The  construction  for  rabattement  is  identical  with  that 
of  Fig.  42,  and  the  point  when  rabatted  appears  in  the  H.P.  at  p2  or  at  pi.  It 
-hould  be  noticed  that  whereas  in  Fig.  42  the  true  inclination  between  the  two 
trace  lines  RS  and  ST  is  a  right  angle,  in  Fig.  43  the  true  inclination  between 
the  traces  of  the  plane  is  the  angle  between  ST  and  a  line  from  S  through  />3  or 
through  p2.  This  angle  may  be  greater  or  less  than  a  right  angle  according  to 
the  arrangement  of  the  traces  of  the  oblique  plane  given. 

In  Fig.  44  is  shown  the  rabattement  of  a  plane  carrying  with  it  a  line  which 
has  previously  been  placed  in  the  plane  and  arranged  so  as  to  be  at  30°  to  the 
H.P.     To  arrange  the  line  in  the  plane  it  must  first  be  placed  in  the  V.P.  with 


Fig.  43. 


one  end  of  it  at  any  convenient  point  hh' ,  the  other  end  of  it  in  A'l'  at  c.  The  plan 
he  is  then  moved,  by  an  arc  with  h  as  centre,  to  bring  the  lower  end  of  the  line 
into  the  H.T.  of  the  plane  at  a.  The  projections  of  the  line  are  ah,  a'b\  and  its 
rabattement  is  at  ah-i  or  at  ah?.. 

An  illustration  of  the  use  of  this  method  is  given  in  Fig.  45,  where  it  is  required 
to  find  the  plan  of  a  square  when  it  is  inclined  to  the  H.P.  at  45°  and  one  edge 
of  it  is  inclined  to  the  same  projection  plane  at  20°.  On  the  rabattement  ahi  of 
the  line  AB  the  figure,  true  size  and  shape,  is  constructed  at  c-zdzCi-  Any  ix)int 
in  this  rabattement  figure  will  move  across  the  H.T.  at  right  angles  until  it 
reaches  its  plan  position.  Thus  c-z  and  d-^  will  be  carried  across  to  their  plans 
c  and  (/  respectively. 

In  order  to  get  the  plan  of  E  from  Ci,  the  line  f.x/j  is  produced  to  the  H.T. 
or  hinge  line  at/,  an  immovable  point,  from  which  a  line  through  d  is  drawn  until 
a  point  opposite  c-z  is  found  at  c.  Parallels  to  cd  and  dc  will  complete  the  plan 
required  of  the  square. 


42 


DESCRIPTIVE   GEOMETRY 


When  the  figure  is  required  to  be  found  in  an  obhque  plane,  then  the  elevation 
obtainable  will  be  a  figure  as  at  Fig.  46.  In  Fig.  45  since  the  plan  only  was  called 
for,  the  plane  of  the  figure,  RST,  was  arranged  perpendicular  to  the  V.P.,  and 
this  would  give  a  line  only  for  elevation,  if  it  were  needed. 

In  Fig.  46  the  plan  and  elevation  are  obtained  of  a  regular  pentagon  placed 
in  a  given  oblique  plane  RST  and  having  one  of  its  edges  at  a  given  angle,  say  20°, 
to  the  H.P.  The  line  ab,  a'b'  is  first  found  in  place,  as  in  the  case  of  Fig.  45,  and 
then  rabatted  over  to  a2&  (see  Fig.  43).  On  this  latter  is  constructed  the  pentagon. 
In  constructing  this  pentagon,  the  angles  of  it  at  c  and  d  are  drawn,  each  108°, 
and  de  and  eg  made  equal  to  cd.  The  fifth  corner  of  the  pentagon  is  then  obtained 
by   intersecting   diagonals   from  c   and  d  made   parallel  to  the  sides   de  and  eg 


Fig.  46. 


respectively.  To  obtain  the  plan  of  the  edge  whose  rabattement  is  de,  produce  de 
to  the  H.T.  at  /  and  from  that  point  draw  a  line  through  the  plan  of  D  till  a  point 
opposite  e  is  found.  For  the  remaining  two  corners  of  the  plan  draw  diagonal 
lines  parallel  to  the  plans  of  the  edges  already  found.  The  elevation  of  the  point 
E  may  be  obtained  by  drawing  a  line  through  the  elevation  of  D  from  the  elevation 
in  XY  of  the  point  F. 

The  use  of  rabattement  is  again  illustrated  in  Fig.  47,  where  two  lines  AB 
and  BC  enclose  an  angle  the  size  of  which  it  is  required  to  find.  As  any  two  lines 
which  meet  each  other  have  a  common  plane  whose  traces  will  contain  the  traces 
of  the  lines,  it  is  only  necessary  to  find  the  H.T.  of  each  of  the  hnes  at  R  and  S 
respectively,  and  the  fine  passing  through  these  is  the  H.T.  of  the  plane  of  the 
lines.  Use  this  H.T.  as  a  hinge  line  about  which  to  rabat  the  plane.  Set  up 
the  usual  right-angled  triangle  whose  hypotenuse  can  be  used  to  swing  down  B 
to  its  rabattement  62.     This  point  joined  to  the  traces  of  the  lines  will  give  the 


RABATTEMENT   OF    PLANES 


43 


rabattement  of  the  angle,  showing  its  true  size.      This  is  marked  in  the  figure 
by  the  arrowed  arc. 


EXERCISK  X\II 

1.  Find  the  plan  of  a  regular  pentagon,  i^"  edge,  whose  plane  is  at  60'  to  the  H.P.  and  which 
has  an  edge  at  30°  to  the  H.P. 

2.  I'"ind  the  plan  and  elevation  of  a  regular  hexagon,  i"  edge,  when  it  lies  in  an  oblique 
plane  with  \'.T.  at  60°  to  XY  and  H.T.  at  45°  to  XY,  and  when  one  edge  of  the  tigure  is  at 
20°  to  the  H.P. 


3.  Find  the  true  inclinations  between  the  lines  meeting  as  at  .1,  B  and  C. 

4.  I'ind  the  true  inclination  between  the  H.T.  and  the  \'.T.  of  any  given  oblique  plar 


By  experimenting  with  two  sticks  of  pencil,  realize  that  a  line,  inclined  to  a 
plane,  makes  an  angle  with  it  which  is  the  complement  of  the  angle  the  same  line 
makes  with  a  perpendicular  to  the  plane  from  any  point  in  the  inclined  line. 

Thus,  in  Fig.  48  the  line  AB  is  evidently  inclined  to  the  plane  RST.  If  a 
perpendicular  to  the  plane  RST  be  drawn  from  A  by  making  its  elevation  per- 
pendicular to  RS  and  its  plan  perpendicular  to  ST,  there  will  be  contained  by 


44 


DESCRIPTIVE   GEOMETRY 


these  two  lines,  AB  and  AC,  an  angle  which  is  the  complement  of  the  angle  which 
AB  makes  with  RST.  Find  this  angle  by  the  method  employed  in  Fig.  47.  The 
angle  indicated  by  the  arrow-headed  arc  in  Fig.  48  is  the  required  angle  the  Hne 
AB  makes  with  the  oblique  plane  RST. 


Fig.  48. 


EXERCISE  XVIII 


Find  the  inclination  to  the  given  oblique  plane,  of  the  line  AB  in  each  of  the  cases  shown 
a.t  A,  B  and  C. 


POINTS   AND    LINES    IN    RELATION   TO    OBLIQUE   PLANES     45 


POINTS  AND  LINES  IX  RKLATION'  TO  OBLIQUE  PLANES 

Sectiox  14.  Realize  by  experiment  the  following  facts: — 
(i)  Parallel  lines  must  be  projected  as  parallel  lines,  i.e.,  their  plans  will  be 
parallel,  and  their  elevations  will  be  parallel,  to  each  other. 

(2)  A  horizontal  line  on  an  inclined  plane  must  be  parallel  to  any  other  hori- 
zontal line  on  the  same  plane  and  therefore  parallel  to  the  H.T.  of  that  plane. 

(3)  A  line  on  an  oblique  plane  must  have  its  V.T.,  if  it  has  one,  in  the  V.T. 
of  that  oblique  plane. 

(4)  A  line  on  an  oblique  plane  and  parallel  to  the  V.P.  is  parallel  to  the  \'.T. 
of  that  plane. 

(5)  A  line  inclined  to  both  planes  of  projection  will  have  its  traces  in  the  traces 
of  any  plane  containing  it. 


Fig.  51. 


Thus  the  plan  ab  in  Fig.  49  is  made  parallel  to  ST  to  represent  a  horizontal 
line  in  the  plane  of  which  ST  is  the  H.T.,  and  when  AB  is  produced  it  will  meet 
the  y.V.  in  a  point  cc'  in  the  V.T.  of  the  plane  RST  in  which  it  Hes.  and  the  ele- 
vation a'b'  of  the  horizontal  line  can  then  be  determined.  This  consideration 
helps  us  to  fmd  projections  of  definite  points  on  oblique  planes.  For  example. 
in  Fig.  50  a  point  on  the  plane  LMX  has  its  plan  given  at  a,  and  there  arc  shown 
three  different  ways  of  obtaining  its  elevation: — i,  by  a  horizontal  line;  2,  by  a  line 
parallel  to  the  V.P.,  and  3,  hy  an  oblique  line,  the  plan  of  the  line  in  each  case 
passing  through  the  point,  and  the  elevation  of  the  line  being  obtained. 

In  Fig.  51  is  shown  how  to  locate,  in  plan  and  elevation,  any  point  on  a  given 
plane  and  having  given  distances  from  the  planes  of  projection,  viz.,  say  15"  from 


46 


DESCRIPTIVE   GEOMETRY 


the  V.P.  and  i"  above  the  H.P.     The  solution  lines  may  be  drawn  in  the  order 
indicated,  resulting  in  first  the  elevation  and  then  the  plan. 

EXERCISE  XIX 

I.  Find  the  other  projection,  in  each  <     -e,  for  the  point  in  the  given  plane,  one  projection 
of  the  point  being  given. 


2.  Draw  traces  for  any  plane,  and  find  the  plan  and  elevation  of  a  point  on  it  f "  in  front  of 
the  V.P.  and  il"  above  the  H.P. 


In  Fig.  52  is  shown  how  to  set  up,  in  plan  and  elevation,  a  perpendicular  to  a 
given  plane  from  any  point  in  it.  It  will  be  remembered  that  a  perpendicular 
to  a  plane  has  its  projections  perpendicular  to  the  traces  of  the  plane.  Let  aa'  be 
the  point.     First  make  a  perpendicular  of  any  convenient  length,  hmited  in  the 


Fig.  52. 

figure  at  the  arrow-head.  Then  turn  this  line  so  as  to  arrange  it  parallel  to  one 
of  the  planes  of  projection,  thus  showing  its  true  length.  Cut  off  the  required 
part  on  this  true-length  line,  and  by  a  parallel  to  XY  the  length  of  one  of  the  pro- 
jections is  determined.     A  perpendicular  across  XY  will  define  the  other  projection. 


POINTS   AND   LINES   IN   RELATION   TO   OBLIQUE   PLANES    47 


EXERCISE  XX 

1.  Find  the  projections  of  a  2"  line  perpendicular  to  a  given  plane  and  starting  from  it  in 
point  i"  from  both  planes  of  projection.  Let  the  plane  be  inclined  at  40°  to  the  H.P.  and  65° 
to  the  V.P. 

2.  The  traces  of  a  plane  are  V.T.  at  30°  to  XV  and  H.T.  at  45°  to  AT.  A  line  perpen- 
dicular to  this  plane  has  a  plan  2"  long.     Find  the  projections  of  a  2"  portion  of  the  line. 


3.  Work  out  on  a  larger  scale  the  following  cases: — 

i.  Find  plan  and  elevation  of  a  point  on  the  H.T.  of  this  plane,  2"  from  the  AT, 

and  set  up  a  perpendicular  line  from  it,  2"  long, 
ii.  Find  a  point  on  this  plane  i\"  above  the  H.P.  and  i"  in  front  of  the  \".P. 
iii.  Find  a  2"  line  perpendicular  to  this  plane  from  the  point  C  in  it. 
iv.  Find  the  elevation  of  AB  and  its  true  length.     It  lies  in  the  given  plane. 
V.  Find  the  plan  of  D  which  lies  in  the  given  oblique  plane.     Given  d'. 
vi.  Find  the  plan  of  KF  which  is  perpendicular  to  the  given  plane,  and  obtain  its 

true  length, 
vii.  Find  the  plan  of  the  line  (HI  which  lies  in  the  given  oblique  plane, 
viii.  Find  the  elevation  of  the  line  JK  which  is  in  the  given  plane,  and  show  its  inclina- 
tions to  the  H.P. 
ix.  The  plan  is  given  of  a  triangle  lying  in  the  oblique  plane  RST.     Find  its  eleva- 
tion then  by  rabattement  of  each  corner  in  turn,  on  to  the  H.P.,  show  the 
true  shape  and  size  of  the  triangle. 

4.  Find  the  true  inclination  between  the  traces  of  each  of  the  planes  at  iii.  iv  and  v. 


48 


DESCRIPTIVE  GEOMETRY 


INTERSECTION  OF  OBLIQUE  PLANES  WITH  EACH  OTHER,  AND  OF  LINES 
WITH  OBLIQUE  PLANES 

Section  15,  If  any  free  line  be  taken,  represented,  for  instance,  by  a  pencil 
for  purposes  of  experiment,  it  will  be  realized  that  any  number  of  planes  may 
contain  it,  and  the  condition  of  this  will  be,  that  the  trace  lines  of  such  planes 
pass  through  or  contain  the  trace  points  of  the  line. 

In  the  illustration  in  Fig.  53  there  are  six  such  planes  shown.  Two  of  them 
are  specially  important  and  useful,  owing  to  the  fact  that  they  can  be  drawn  or 


X  y 


yV 


:^'   I     3 


i\ 


!        .1 


]^-^^4i.T. 


Fig.  53. 


represented  without  the  necessity  of  first  discovering  the  trace  points  of  the  given 
line.  They  are  Nos.  2  and  4 — the  one,  a  plane  perpendicular  to  the  V.P.,  and 
the  other  a  vertical  plane  or  plane  perpendicular  to  the  H.P. 

The  next  thing  to  realize  is  that  since  planes  intersect  each  other  in  straight 
lines,  it  will  easily  be  possible  to  show  the  plan  and  elevation  of  an  intersection 
line  by  joining  the  point  on  the  V.P.  common  to  the  Vertical  Traces  of  the  two 
planes  (i.e.,  where  the  two  V.T.'s  meet),  to  the  point  on  the  H.P.  where  the  planes' 
H.T.'s  meet.  In  Fig.  54  an  illustration  is  given  where  plane  RST  intersects  plane 
LMN  in  line  AB.  The  plan  of  the  intersection  line  is  at  ah  and  its  elevation 
is  a'h'. 


INTERSECTION  OF  OBLIQUE  PLANES  WITH  EACH  OTHER    49 

Further  realize  that  a  free  Unc  directed  to  any  plane,  or  passing  through  it, 
will  h^e  an  intersection  point  on  that  plane,  and  this  will  be  a  po.nt  ,n  the  mter- 


Fig.  54. 


rzr:ri:T.t=:i:n/:r;rL='r.v" 


Fio.  55- 


A  plane  /,.1/.V  containing  AB  intersects  ^r  in  the  line  passing  through  V  and  H 
and  shown  in  plan  and  elevati.in  at  vH  and  I  .1/. 


50 


DESCRIPTIVE   GEOMETRY 


When  the  plan  of  the  line  AB,  namely,  ab,  is  produced  to  meet  the  plan  of 
this  intersection  line,  namely,  vH,  the  plan  of  the  intersection  point  is  found  at 
P,  and  P',  the  elevation  of  it,  can  be  at  once  obtained. 

Be  careful  to  note  that  if  a  vertical  plane  is  used  in  which  to  contain  the  given 
line,  then  the  elevation  of  the  intersection  will  meet,  or  cross,  the  elevation  of  the 
line,  or  the  line  produced,  at  the  elevation  of  the  intersection  point  required,  and 
that  if  a  plane  perpendicular  to  the  V.P.  is  used  to  contain  the  line,  it  will  be  the 
plan  of  the  intersection  line  that  will  meet  the  plan'  of  the  given  line  in  the  plan 
of  the  intersection  point  required. 


EXERCISE  XXI 

I.  Find  the  intersection,  showing  it  by  plan  and  elevation,  of  the  line  AB  with  the  plane 
/?5r  in  each  of  the  cases  ^,  5  and  C. 


Y         1 


2.  Find  the  distance  of  the  point  A  from  the  plane  RST  at  D. 

3.  Find  the  projections  of  the  intersection  line  in  each  of  the  cases  at  E,  F,  G  and  H  where 
planes  are  arranged  so  as  to  intersect  each  other.     Work  to  a  large  scale. 


CHAPTER  VIII 
PARALLEL  PLANES 

Section  16.  It  needs  no  demonstration  to  realize  that  planes  parallel  to  one' 
another  will  meet  the  planes  of  projection  in  parallel  trace  lines. 

In  Fig.  56  at  (i)  two  vertical  planes  are  shown,  and  it  is  evident  that  the  dis- 
tance between  these  planes  is  not  the  distance  between  their  Vertical  Traces, 
which  may  be,  for  these  planes,  a  varying  distance  apart  according  to  the  angle 


Fic.  56. 


at  which  they  meet  the  V.P.,  but  the  distance  between  them  is  the  distance  between 
their  Horizontal  Traces,  because  they  meet  the  H.P.  perpendicularly. 

In  the  case  at  (ii)  the  distance  between  the  planes  is  the  same  as  that  between 
their  V.T.'s,  because  in  this  case  the  planes  meet  the  V.P.  perpendicularly. 

At  (iii)  the  parallel  planes  are  oblique,  and  the  method  of  Imding  the  distance 
between  them  is  to  start  a  jcrpendicular  line  from  som£_4iiiLnt  in  one  of  them, 
say,  PQZ.  and  find  where  it  intersects  the  other  plane  RST,  then  fmd  the  length 
of  the  line  from  one  to  the  other.  To  do  this,  take  any  point  in  the  H.T.  of  the 
plane  PQZ,  as  at  aa',  and  from  it  set  up  a  perpendicular,  marked  with  arrow-heads 
in  the  figure.     Now  consider  this  line  to  be  in  a  vertical  plane  L.l/.V,  and  find  the 


52 


DESCRIPTIVE  GEOMETRY 


intersection  of  LMN  with  RST,  giving  a  line  whose  elevation  is  VH  and  which 
contains  point  P  shown  at  p'p.  Find  the  true  length  of  AP  at  a'p'2'  This  is  the 
distance  between  the  two  given  planes. 


EXERCISE  XXII 

Find  the  true  distance  between  the  parallel  planes  A  and  B,  also  between  C  and  D. 


It  was  previously  seen  that  a  horizontal  line  in  a  plane  was  parallel  to  the 
H.T.  of  that  plane,  and'  now  it  will  be  seen  that  any  horizontal  line  parallel  to  a 
plane  is  parallel  to  the  H.T.  of  that  plane.  And,  as  parallel  lines  have  plans  parallel 
to  one  another,  it  will  be  realized  that  when  two  planes  are  parallel  to  one  another, 


Fig.  57. 


the  H.T.  of  one  of  the  planes  is  parallel  to  any  horizontal  line  on  the  other  plane. 
Hence  the  solution  given  in  Fig,  57  for  finding  a  plane  parallel  to  ?i  given  on^^ 
and  at  a  given  distance  from  it. 


PARALLEL   PLANES 


53 


From  any  point  A  in  the  given  plane  RST  set  up  a  perpendicular  AB  and  cut 
ofT  a  part,  AP,  say,  i"  long.  Through  P  draw  a  horizontal  hne  parallel  to  the 
plane  RST  by  making  its  plan  pq  parallel  to  ST  and  its  elevation  p'q'  parallel  to 
XY.  The  V.T.  of  this  line  is  at  q\  which  is  also  a  point  in  the  V.T.  of  the  required 
parallel  plane.  Draw  LM  parallel  to  RS,  and  3/-V  parallel  to  ST.  It  should 
be  noticed  that  the  plane  LMX  has  its  traces  perpendicular  to  the  projections  of 
the  line  AB,  and  that  this  plane  has  been  passed  through  a  certain  definite  point 
P,  marked  in  the  line  AB. 

So,  likewise,  if  it  be  required  to  pass  a  plane  through  any  given  point  P,  Fig.  58, 
and  to  arrange  it  perpendicularly  to  any  given  line,  AB,  it  is  only  necessary  to 
make  a  second  hne,  CP,  through  P,  with  its  plan  perpendicular  to  the  plan  of  the 
given  line  AB,  and  to  represent  a  horizontal  line,  in  this  way,  on  the  required 
plane.  This  line  CP  will  have  its  V.T.  at  V,  which  is  a  point  in  the  V.T.  of  the 
required  plane  RST.  RS  will,  of  course,  be  made  perpendicular  to  the  elevation 
a'b'  of  the  given  line. 


A  problem  illustrating  the  application  of  two  or  three  recently  discussed 
methods  is  worked  out  in  Fig.  59,  where  it  is  required  to  lind  the  centre  of  the 
sphere,  which  has  upon  its  surface  four  points  whose  projections  are  given.  The 
four  points  are  shown  in  plan  and  elevation  at  aa,  bb',  cc'  and  dd' .  Any  point 
on  the  surface  of  a  sphere,  joined  to  any  other  point  on  the  surface  by  a  straight 
line,  will  give  a  chord  of  the  sphere,  and  a  plane  bisecting  this  chord  will  cut  the 
sphere  into  two  equal  parts,  that  is,  will  pass  through  the  centre  of  the  sphere. 
Notice  also  another  fact,  namely,  that  if  three  points  only  be  chosen  as  points  on 
the  surface  of  a  sphere,  the  size  of  the  sphere,  upon  the  surface  of  which  these 
points  may  lie,  may  be  any  size  provided  that  it  is  not  less  in  diameter  than  the 
circle  upon  the  circumference  of  which  these  points  may  have  place.  Con- 
sequently, in  order  to  limit  the  size  of  the  sphere  to  be  dealt  with,  a  fourth  point  is 
necessary,  and  all  four  of  the  points  must  be  made  use  of  in  the  solution. 

By  joining  the  given  points,  several  chords  may  be  obtained,  as  shown  in  plan 
and  elevation.  Fig.  59,  and,  according  to  the  argument  above,  if  planes  are  made 


54 


DESCRIPTIVE   GEOMETRY 


to  bisect  them  at  right  angles,  each  plane  so  found  will  pass  through  the  centre 
of  the  sphere.  Thus,  by  joining  ab  and  a'b'  the  chord  AB  \s  represented,  and 
from  the  centre  point  of  it,  pp',  a  horizontal  line  PQ  is  drawn  as  a  horizontal  line 
on  the  plane  perpendicular  to  AB,  giving  the  point  q'  as  its  V.T.  through  which 
the  V.T.  of  the  AB  plane  may  be  drawn  perpendicular  to  a'h'.  Its  H.T.  is  then 
drawn  perpendicular  to  ah.  So  again  in  the  two  other  cases,  giving  the  AC  plane 
and  the  CD  plane.     The  plan  and  elevation  of  the  intersection  of  plane  CD  with 


Fig.  59. 


plane  ^C  is  next  found,  and  then  the  intersection  of  plane  AB  with  one  of  these 
(conveniently  AC  m  this. solution),  and  the  plans  of  the  intersections  obtained 
cross  each  other  in  the  plan  of  the  centre  of  the  sphere.  Likewise  the  elevations 
of  the  intersections  cross  each  other  in  the  elevation  of  the  centre  of  the  sphere. 

The  radius  of  the  sphere  may  now  be  found,  if  desired,  by  obtaining  the 
true  length  of  the  line  joining  this  centre  point  to  any  of  the  given  points  A,  B,  C 
or  D.  The  sphere  may  then  be  represented  by  circles,  one  for  plan  and  one  for 
elevation,  with  the  points  found,  as  centres. 


PARALLEL   PLANES 


55 


EXERCISE  XXIII 

I.  Find  the  traces  of  a  plane  parallel  to,  and  i^"  distant  from,  a  given  plane  shown  at  A, 
whose  V.T.  is  at  60°  to  XY  and  H.T.  at  30°  to  XV. 


2.  Find  the  traces  of  a  plane  parallel  to  the  given  one  shown  at  B  and  i"  perpendicular 
distance  from  it. 

3.  Kind  a  plane  perpendicular  to  the  given  line  shown  at  C.  and  passing  through  the  given 
point  P. 


T/ 
1            »("                                                       I 

!        1                                     '          »»' 

\             -       1        1                                     11 

1           1                                             T^'      !                i 

•«'            j       !                             Ill 

1                        lit'''                                      '1                '      t/'' 
A'           '                    II'                                    !         1              1     1                      V 

k  i                          '1 

j                                               |v    id                                                     it.           I                              1 

1        •              1  \                                1              j 

1           1                  1       ''* 

id                   1  •                                 ! 

4.  Find  the  centre  of  the  sphere  which  has  on  its  surface  the  four  given  points  shown  at 
A  BCD. 

5  Find  the  plan  and  elevation  of  the  sphere  which  has  on  its  surface  the  four  given  points 
EFGU 


56 


DESCRIPTIVE  GEOMETRY 


DIHEDRAL  ANGLES  CONTAINED  BY  OBLIQUE  INTERSECTING  PLANES, 
AND  PLANES  DIVIDING  THESE  ANGLES 

Section  17.  In  considering  dihedral  angles,  or  angles  contained  by  planes 
as  they  incline  to  one  another,  the  illustration  at  (i),  Fig.  60,  shows  two  planes 
RST  and  LMN  meeting  each  other  in  a  vertical  line  at  H,  and  meeting  the  H.P. 
at  right  angles  with  the  result  that  the  angle  and  its  supplement,  named  the  dihedral 
angles  contained  by  these  two  planes,  are  equalled  by  the  angles  MHS  and  SEN 
on  the  H.P. 

Similarly,  at  (ii),  the  two  planes  RST  and  LMN  meet  the  V.P.  at  right  angles, 
and  the  dihedral  angles  contained  by  them  are  evident  at  MVS  and  SVL,  because 


Fig.  60. 


they  are  equalled  by  the  rectilineal  angles  contained  by  the  traces  or  intersections 
RS  and  LM  on  the  V.P.,  that  is,  on  the  plane  perpendicular  to  them. 

At  (iii)  the  planes  are  so  arranged  that  they  are  not  perpendicular  to  one 
of  the  planes  of  projection,  and  consequently  a  third  plane,  other  than  one  of  the 
planes  of  projection  which  served  the  purpose  in  cases  (i)  and  (ii),  must  be  made 
use  of,  in  order  that  the  intersections  by  the  two  given  planes  made  on  this  third 
plane  may  be  found,  and  the  angles  by  the  intersections  on  it  measured. 

Because  at  (iii)  the  planes  are  so  arranged  as  to  intersect  each  other  in  a  hori- 
zontal line  not  perpendicular  to  the  vertical  plane  of  projection,  the  third  plane, 
with  its  H.T.  marked  3,  is  arranged  so  that  it  cuts  the  given  planes  at  right  angles. 
It  cuts  the  intersection  line  of  the  given  planes  at  P,  and  when  this  point  is  rabatted 
onto  the  H.P.  at  P2  and  joined  to  a  and  b  where  the  third  plane's  H.T.  crosses 
the  H.T.'s  of  the  given  planes,  the  angle  aP^b  and  its  supplement  are  obtained, 
to  which  the  dihedral  angles  required  are  equal. 


DIHEDRAL   ANGLES-OBLIQUE  INTERSECTING  PLANES      57 

In  Fig.  6 1  the  two  planes  are  so  arranged  that  their  common  intersection 
line  is  inclined  from  the  point  v'  where  the  vertical  traces  cross,  to  the  point  H 
where  the  horizontal  traces  cross.  In  this  case  a  third  plane,  perpendicular  to  the 
two  given  ones,  will  therefore  be  inclined,  and,  being  at  right  angles  to  the  intersec- 
tion line  VH,  its  H.T.  marked  3,  must,  of  course,  be  made  at  right  angles  to  the 
plan,  vH,  of  the  intersection  line. 

By  turning  vH,  with  centre  v,  into  the  XF,  and  carrying  with  it  the  point  c, 
the  real  inclinations  of  the  intersection  line  and  of  the  third  plane  may  be  repre- 
sented on  the  V.P.,  where  the  line  I'P  is  a  view  of  the  intersection  line  showing 


»\ 

r 

^ 

/ 

X    / 

X 

.u\ 

.y 

'^^^\. 

\^ 

\ 

/ 

h 

1 
1 
1 

x*^ 

XT/ 

7  ^^ 
/    ^ 

1 

1 

Fig.  61. 


its  inclination  to  the  H.P.  and  the  line  at  right  angles  to  it  through  P  shows  the 
inclination  of  the  plane  3.  Point  P  marks  the  level  at  which  they  intersect.  From 
the  point  P,  where  the  inclined  intersection  line  passes  through  the  inclined  plane  3, 
the  distance,  indicated  by  a  bracket,  down  the  plane  3  to  the  H.P.  is  then  taken 
and  transferred  to  cPi.  Join  Pi  to  a  and  to  h,  and  these  lines,  which  are  the 
rabattements  of  intersections  made  by  RST  and  LM'S ,  respectively,  with  plane 
3,  as  in  case  (iii),  Fig.  60,  will  give  angles  to  which  the  dihedral  angles  required 
are  equal,  namely,  aP-ih  and  its  supplement. 

In  working  exercises,  it  is  advisable,  until  the  student  is  familiar  with  each 
step,  to  make  hair  lines  and  attach  names  to  all  the  various  lines  and  points  whi!<^ 
the  solution  is  in  progress. 


58 


DESCRIPTIVE   GEOMETRY 


(Notice  that  the  third  plane's  H.T.,  marked  3,  and  arranged  at  any  convenient 
place  perpendicular  to  the  plan  vH  of  the  intersection  line,  may  need  to  be  produced 
beyond  XY  in  order  to  meet  the  H.T.  of  LMN  at  h,  as  in  Fig.  62.) 


EXERCISE  XXIV 

Find  and  mark  the  angles  contained  by  the  planes  represented  in  pairs  at  (i),  (2),  (3)  and 
(4).     Make  the  drawings  very  much  larger  than  what  is  shown  here. 


By  experiment  with  an  open  book,  it  will  be  realized  that  any  dihedral  angle, 
contained  by  the  covers  of  the  book,  may  be  divided  by  arranging  a  leaf  or  leaves 
of  the  book,  so  that  from  one  cover  to  a  leaf  is  a  dihedral  angle,  and  from  that 
leaf  to  another,  separated  from  it,  or  to  the  other  cover,  will  be  another  dihedral 
angle.  The  leaf,  representing  a  plane,  divides  the  dihedral  angle  between  the 
covers  which  represent  other  planes,  and  will  be  seen  to  have  the  same  intersec- 
tion line  as  the  covers  have,  namely,  the  hinge  edge  of  the  book. 

Now,  let  it  be  required  to  find  the  traces  of  planes  which  will  bisect  the 
dihedral  angles  contained  by  any  two  planes  whose  traces  are  given.  It  will  be 
clear  that  such  traces  will  pass  through  the  traces  of  the  intersection  line  made  by 
the  given  planes,  since  they  must  contain  the  same  intersection  line. 

In  Fig.  62  the  planes  RST  and  LMN  are  the  given  planes  with  their  inter- 
section Hne  passing  through  V  and  H.  After  proceeding  as  in  the  case  explained 
in  Fig.  61,  the  rabatted  intersections  oi  RST  and  LMN  with  plane  3  will  give  the 
angles  aP2b  and  aPod  marked  by  arrow-headed  arcs,  showing  the  sizes  of  the 
dihedral  angles  contained  by  the  given  planes. 

Now,  bisectors  of  these  angles,  marked  in  the  figure  by  straight  lines  with  arrow- 
heads, will  serve  as  rabatted  intersections  made  with  plane  3  by  planes  bisecting 
the  dihedral  angles,  and,  of  course,  these  intersection  lines  will  have  their  H.T.'s 
in  the  H.T.  of  plane  3  at  e  and/  respectively,  and  also  it  will  be  realized  that  these 


DIHEDRAL   AXGLES-OBLIQUE   INTERSECTING   PLANES 


59 


H.T.'s  will  be  in  the  H.T.'s  of  the  required  bisecting  planes.  Since  the  bisecting 
planes  contain  the  intersection  line  passing  through  V  and  //,  these  points,  V  and 
//,  are  in  the  traces  of  the  bisecting  planes  required.  The  result  is.  that  the  line 
through  II  and  c  is  the  H.T.,  and  from  where  this  meets  XV  a  line  through  -/  is 
the  V.T.,  of  one  of  the  required  planes,  while  ///  is  the  H.T.,  and  a  line  through 
v'  to  meet  JH  in  XY,  is  the  V.T.  of  the  other  required  plane. 

If,  as  is  the  case  in  this  figure,  there  is  not  room  enough  to  lind  the  meeting 
of  the  H.T.  and  V.T.  on  XY  for  the  last  plane,  then  a  point  xx'  should  be  chosen 


Fig.  62. 


on  the  common  intersection  line,  and  a  horizontal  line  on  the  required  plane  should 
be  drawn  to  meet  the  \'.P.  at  kk'.  This  latter  point  on  the  \'.P.  is  in  the  V.T. 
required,  which  is  obtained  by  joining  v'  to  k'. 

If  the  H.T.  of  this  last  plane,  marked /F,  is  parallel  to  AT,  of  course  the  \'.T. 
of  it,  passing  through  ?'  \\i\\  also  be  parallel  to  A'l^. 

Also,  if  ^he  bisector  Pyf  runs  nearly  parallel  to  H.T.  3.  so  that  there  is  not 
room  enough  to  find/,  then  the  direction  of  the  line///  may  be  obtained  by  making 
use  of  an  inclined  line  in  the  plane  required. 


60 


DESCRIPTIVE   GEOMETRY 


EXERCISE  XXV 

Find  the  traces  of  planes  which  bisect  the  dihedral  angles  contained  by  the  pairs  of  planes 
given  at  A  and  at  B. 


It  will  now  be  seen,  that,  by  supposing  the  rabattement  of  the  angle  between 
two  planes,  and  working  conversely  to  the  method  in  Fig.  6i,  it  will  be  possible 
to  find  the  traces  of  a  second  plane  at  a  given  angle  to  a  first  one,  whose  traces 
may  be  given,  and  intersecting  it  in  a  line  of  given  inclination  to  the  H.P. 


Fig.  63. 


For  illustration,  let  the  plane  RST,  Fig.  63,  be  given,  and  let  it  be  required  j 
to  find  the  traces  of  another  which  will  cut  into  this  one  in  a  line  of  30°  to  the! 
H.P.,  and  make  a  dihedral  angle  of,  say,  70°  with  this  given  plane. 


DIHEDRAL    ANGLES-OBLIQUE   INTERSECTING  PLANES     61 

From  any  point  V  in  RS  draw  a  line  on  the  V.P.  at  30^  to  XY.  With  :•,  the 
plan  of  V,  as  centre,  bring  the  plan  of  this  line  around  by  an  arc  until  its  lower  end 
is  at  //  in  ST.  Then,  at  any  convenient  place,  cross  this  plan  of  the  intersection 
line  by  the  H.T.  of  plane  3,  previously  made  use  of  in  Figs.  60  and  61,  and  carry 
the  point  c  into  XY  so  as  to  show  from  the  point  there  obtained  the  inclination 
of  plane  3  and  its  intersection  of  the  given  intersection  line  at  P.  The  bracket 
line,  now  measured  off  at  cPj,  will  give  a  i)oint  which  when  joined  to  a  will  show 
the  rabattement  of  intersection  of  the  given  plane  RST  with  the  plane  3.  Apply 
to  this  rabattement  the  angle  70°  and  so  obtain  the  rabatted  intersection  Pzb  of 
the  required  plane  with  plane  3,  and  b,  a  point  in  its  H.T. 

By  joining  H  and  b  and  producing  the  line  to  XY  the  H.T.  of  the  required 
plane  is  found,  and  its  V.T.  may  then  be  found  by  drawing  a  line,  from  this  point 
in  XY,  through  v'. 


EXERCISE  XXVI 


Find  the  traces  of  a  plane  which  makes  with  the  given  plane  an  angle  of  65°  and  intersects 
it  in  a  line  inclined  at  30°  to  the  H.P.     Two  cases,  .4  and  B. 


CHAPTER  IX 

PROJECTION  OF  RECTILINEAL  ANGLES,  AND  OF  FIGURES  AND  SOLIDS 
INVOLVING  THE  SAME 

Section  18.  Two  inclined  lines  meeting  each  other  at  a  point  above  the  H.P. 
may  be  made  to  form  any  angle  as  they  meet,  and  if,  say,  two  sticks  of  pencil, 
to  serve  for  lines,  be  used  experimentally,  it  may  be  seen  that  the  projections  of 
the  two  lines  and  of  the  angle  they  contain  will  vary  according  to  how  the  pair 
of  lines  is  disposed  in  relation  to  the  planes  of  projection.     Notice  that  the  plane 


Fig.  64. 


of  the  two  lines,  that  is,  the  plane  in  which  both  have  place,  varies  in  its  attitude, 
with  the  variation  of  the  angle  formed  by  the  two  lines,  while  the  lines  may  still 
remain  each  at  its  own  particular  angle  to  the  H.P. 

As  illustration,  let  there  be  two  lines  A  and  B.  Let  A  be  inclined  at,  say, 
35°  to  the  H.P.,  and  B  have  an  inclination  of,  say,  50°  to  the  H.P.  It  must  first 
be  realized  that  the  greatest  possible  angle  that  can  be  contained  by  these  lines 
is  95°,  and  in  order  that  this  may  be  so,  the  plane  of  the  lines  would  have  to  be 
vertical.  Such  an  arrangement  of  them  is  shown  in  Fig.  64,  where  the  plans  of 
the  lines  are  parts  of  the  XY  line.  Realize  that  the  lengths  of  their  plans  will 
not  vary  so  long  as  the  lines  are  not  increased  in  length  and  their  angles  with  the 
H.P.  are  not  changed.  The  plans,  however,  may  be  moved  the  one  toward  the 
other  so  that  instead  of  the  distance  between  their  lower  ends  or  H.T.'s  being  as 
great  as  from  a  to  ^  in  the  figure,  it  may  be  reduced  to,  say,  the  distance  a  to  b2, 
the  line  B  being  moved  so  as  to  have  its  H.T.  at  bo.     The  angle  contained  by  the 

62 


PROJECTION   OF   RECTILINEAL   ANGLES 


63 


two  lines  A  and  B  is  now  reduced,  since  the  distance  a  to  bo,  subtending  the  angle, 
is  reduced,  but  not  the  lengths  of  the  lines.  The  new  elevation  of  B  is  now  at 
B2,  and  the  angle  contained  by  A  and  B  has  been  reduced  to  an  angle  of  which 
the  plan  and  elevation  are  indicated  by  the  arrow-headed  arcs.  The  true  size 
of  this  angle  can  readily  be  obtained  by  rabattement  about  ab^  as  a  hinge  line,  abz 
being  the  H.T.  of  the  plane  of  the  two  lines  as  now  arranged. 

As  a  specific  case,  let  it  be  required  to  find  plan  and  elevation  for  an  angle 
contained  by  a  pair  of  lines  of  any  length  when  one  of  them  is  at  30°  to  the  H.P. 
and  the  other  is  at  45°  to  the  H.P.,  and  the  angle  contained  by  them  is  80°. 

By  placing  both  lines  in  the  V.P.  as  in  Fig,  65-,  (i),  at  A  and  B,  starting  them 
both  from  the  same  point  and  producing  them  downward  to  their  H.T.'s  at  a  and  b 
respectively,  their  plans,  proper  for  the  particular  angles  they  make  with  the  H.P., 
are  found.  Allowing  the  30°  line  A  to  remain  in  the  V.P.,  the  45°  line  B  must 
be  moved  toward  A  in  order  to  reduce  the  angle  from  what  it  is  at  present  (105°) 


Fic.  65. 


to  80°.  This  80°  will  be  subtended  by  a  line,  across  the  H.P.  between  the  H.T.'s 
of  the  lines,  the  length  of  which  must  be  found,  therefore  with  c'  as  centre  move 
B  toward  A  until  80°  is  enclosed  as  shown  by  the  arrow-headed  arc.  This  gives 
the  distance  abo  required  to  subtend  80°  when  the  lines  are  A  and  B.  Now,  with  c 
as  centre  bring  the  plan  cb  round  by  an  arc  until  the  H.T.  of  B  is  at  its  proper  dis- 
tance from  the  H.T.  of  A.  This  will  require  the  use  of  an  intersecting  arc  with 
a  as  centre  and  (7/)2  as  radius.  The  new  place  for  the  H.T.  of  B  is  ^3-  Join  it  to  c 
for  the  new  position  of  plan,  and  then  obtain  the  elevation  of  the  line  as  shown. 
By  joining  a  to  b^  the  H.T.  of  the  plane  of  the  two  lines  is  obtained,  and  the  V.T. 
of  the  plane  of  the  two  lines  will  be  dc\ 

In  Fig.  65  at  (ii)  is  shown  a  particular  case  which  is  very  useful.  The  case 
is  that  of  the  right  angle  (90°)  contained  by  two  lines,  the  sum  of  whose  angles 
to  the  H.P.  is  less  than  90°.  The  two  lines  are  first  placed  as  at  A  and  B  in  the 
V.P.,  with  their  plans  therefore  in  AT.  If,  now,  A  be  allowed  to  remain  in  the 
V.P.  wath  ca  as  its  plan,  and  B  is  to  be  placed  so  as  to  contain  with  .1  an  angle 


64  DESCRIPTIVE   GEOMETRY 

of  90°,  then  the  elevation  of  B  will  be  a  perpendicular  to  the  elevation  A,  because, 
as  we  have  seen  previously,  when  a  right  angle  is  contained  by  two  lines,  and  one 
of  them  is  in  a  projection  plane  or  parallel  to  it,  then  the  other  has  its  projection 
on  the  same  plane  perpendicular  to  the  projection  of  the  first.  Therefore  c'h'  is 
at  once  obtained  as  the  elevation  of  the  B  line,  and  its  plan  can  then  be  found 
by  making  a  perpendicular  from  h'  and  intersecting  it  by  an  arc  with  c  as  centre, 
which  carries  down  the  plan  length  ch  into  its  new  place.  Notice  that  by  joining 
a  toh  the  H.T.  of  the  plane  of  the  two  lines  is  obtained,  and,  of  course,  the  V.T. 
of  the  same  plane  is  ac',  coinciding  with  the  elevation  ac'  of  the  line  A  which  is 
in  the  V.P. 

N.B. — When  the  sum  of  the  angles  made  with  the  H.P.  by  the  given  Hnes 
is  over  90°,  the  angle  contained  by  them  must  be  less  than  90°.  Consequently, 
for  large  contained  angles  the  inclinations  of  the  lines  to  the  H.P.  must  be  small 
accordingly. 


EXERCISE  XXVII 

1.  Find  the  plan  and  elevation  of  an  angle  of  45°  contained  by  two  lines  when  one  of  them 
is  in  the  V.P.  and  inclined  to  the  H.P.  at  35°,  and  the  other  is  at  50°  to  the  H.P. 

2.  Find  the  plan  and  elevation  of  two  lines,  one  at  25°  to  the  H.P.,  and  in  the  V.P.,  and  the 
other  at  45°  to  the  H.P.,  when  they  contain  an  angle  of  100°. 

3.  Find  the  plan  and  elevation  of  two  lines,  one  at  45°  to   the  H.P.  and  in  the  V.P.,  and 
the  other  at  30°  to  the  H.P.,  when  they  contain  an  angle  of  45°. 

4.  Find  the  plan  and  elevation  of  an  angle  of  90°  when  contained  by  two  lines,  one  at  35° 
to  the  H.P.,  and  the  other  at  20°  to  the  H.P. 

N.B. — In  each  case  mark  which  is  the  plan  of  the  contained  angle,  and  which  is  the  ele- 
vation of  it. 


An  application  of  the  problem  above  discussed  may  now  be  made  to  the  finding 
of  the  projections  of  any  rectilineal  figure,  triangle,  square  or  polygon,  when  two 
adjacent  sides,  or  a  diagonal  and  an  adjacent  side,  or  two  diagonals,  are  given 
at  inclinations  to  the  H.P.  Instead  of  inchnations  for  edges,  etc.,  levels  may 
be  given  for  corners,  and  the  inclinations  of  edges  or  diagonals  ascertained 
accordingly. 

In  Fig.  66  is  represented  in  plan  and  elevation  a  square  having  one  edge  at 
25°  to  the  H.P.  and  in  the  V.P.,  and  an  adjacent  edge  at  40°  to  the  H.P.  As  the 
angle  contained  by  the  two  Unes  is  90°,  the  40°  line  can  be  found  in  its  proper  place 
in  relation  to  the  25°  one,  by  the  method  of  Fig.  65  (ii).  The  edge-length  of  the 
square  must  be  marked  off  as  c'a'  on  the  true  length  of  the  25°  line  and  as  c'h'2  on 
the  true  length  of  the  40°  line,  a'c'  gives  ac  as  its  plan,  and  from  c'h'2  may  be 
obtained  c'h'  and  ch,  the  elevation  and  the  plan  respectively,  of  the  40°  edge  of 
the  square.  Opposite  edges  of  the  figure  being  parallels,  their  projections  may 
be  obtained  by  making  them  parallel  to  those  already  obtained.     The  square 


PROJECTIOX   OF   RECTILINEAL   AXGLES 


65 


thus  found  may  now  be  considered  as  the  face  of  a  cube,  and  it  will  be  seen  that 
by  drawing  a  line  through  ////i,  which  points  are  the  H.T.'s  of  the  two  edges  of 
the  square,  the  H.T.  of  the  plane  of  the  square  is  obtained,  and  Ila'c'  will  be  the 
V.T.  of  the  plane  of  the  square.  A  perpendicular  to  this  plane  from,  say,  the 
point  A  may  be  made  by  the  method  explained  in  Fig.  52.  If  this  perpendicular 
be  made  equal  in  length  to  the  edge  of  the  square,  then  it  will  be  seen  that  the 
cube  can  be  represented  by  drawing  other  perpendiculars  with  projections  equal 
in  length  to  the  projections  of  this  one  and  parallel  to  it;  and  by  joining  the  upper 
ends  of  them  the  drawing  will  be  completed.  To  correctly  represent  this  cube 
as  a  solid  the  three  edges  in  plan  from  the  corner  d  should  be  made  dotted  lines, 
and  the  edge  a'c'  in  the  elevation  should  also  be  made  a  dotted  line. 


Fig.  66. 


In  Fig.  67  the  plan  and  elevation  of  a  regular  pentagon  (whose  angle  is  108°) 
is  found,  one  edge  being  at  25°  to  the  H.P.  and  an  adjacent  edge  at  35°  to  the 
H.P.  Following  the  construction  shown  in  Fig.  65,  (i),  the  inclined  lines  are  found 
in  position  at  ac,  ch  for  plans,  and  ac',  c'b'  for  elevations.  Point  C  is  then  rabatted 
to  Co,  where  the  angle  108°  appears  at  its  true  size,  and  the  figure  is  formed  with 
this  angle  for  one  of  its  corners.  From  the  figure  so  formed  in  rabattement  the 
corner  points  are  carried  over  the  H.T.  line  ab  perpendicularly  to  the  plan,  and 
then  from  the  plan  the  elevation  is  obtained.  Advantage  is  taken  of  the  fact 
that  there  is  a  diagonal  of  the  pentagon  parallel  to  each  side  of  it. 

The  plan  and  elevation  of  the  pentagon  having  been  completed,  the  centre 
of  it  may  be  obtained  by  directing  lines  from  the  centres  of  any  two  sides  to  oppo- 
site corners  and  letting  them  intersect  as  at  dd' . 

The  pentagon  might  be  considered  to  be  the  base  of  a  pyramid,  and  then  a 
perpendicular  from  dd'  may  be  found,  of  a  given  length,  to  serve  as  the  axis  of  the 
solid,  by  the  method  of  Fig.  52.  The  upper  end  of  this  axis,  i.e.,  the  apex  of  the 
pyramid,  may  then  be  joined  by  straight  lines  to  the  corners  of  the  pentagon  to 


66 


DESCRIPTIVE   GEOMETRY 


complete  the  representation  of  the  solid.  Care  must  be  taken  to  make  certain 
lines  dotted  lines  to  represent  them  as  being  out  of  view,  in  the  plan  and  in  the 
elevation.     In  Fig.  67  the  apex  pp'  is  found,  but  the  solid  is  not  shown. 


EXERCISE  XXVIII 

1.  Find  the  plan  and  elevation  of  a  regular  pentagon  i\"  edge,  when  one  edge  is  at  20°  to 
the  H.P.  and  an  adjacent  edge  is  at  35°  to  the  H.P. 

2.  Find  the  plan  and  elevation  of  a  right  hexagonal  pyramid  when  one  edge  of  the  base  is 
at  15°  to  the  H.P.  and  an  adjacent  edge  of  the  base  is  at  40°  to  the  H.P.  The  length  of  t*he 
base  edge  is  i",  and  of  the  axis  3". 

3.  Find  the  plan  and  elevation  of  a  cube  2"  edge,  when  one  edge  is  at  25**  to  the  H.P.  and 
an  adjacent  edge  is  at  50°  to  the  H.P. 

4.  Find  the  plan  and  elevation  of  a  pentagon,  1"  edge,  when  three  consecutive  corners  of 
it  are  at  levels  \" ,  l"  and  if",  above  the  H.P. 


THE  TETRAHEDRON  AND  THE  OCTAHEDRON 


Section  19.  Of  the  "  regular  "  soHds,  we  have  already  dealt  with  the  cube, 
and  we  shall  now  proceed  to  consider  two  others,  the  tetrahedron,  with  four  equal 
faces,  each  an  equilateral  triangle,  and  the  octahedron,  which  has  eight  equal  faces, 
each  being  an  equilateral  triangle. 


THE  TETRAHEDRON  AND  THE  OCTAHEDRON 


67 


The  appearance  and  characteristics  of  these  solids  may  be  judged  from  the 
representations  of  them  in  Fig.  68  at  (i)  and  (ii). 

The  tetrahedron  has  a  corner  perpendicularly  opposite  the  centre  of  each 
face;  while  the  perpendicular  distance  of  a  corner  from  the  opposite  face  is  the 
height  of  the  solid.  The  simplest  way  by  which  to  find* the  height  of  any  tetra- 
hedron is  to  place  it  on  the  H.P.,  as  shown  at  (iii),  and  then,  since  all  of  its  edges 
are  equal,  and  the  edge  ^-B  is  parallel  to  the  V.P.,  a'h'  shows  true  length,  its  plan 
being  at  ah.    The  height  of  the  solid  is  therefore  found  to  be  b'c'. 

The  octahedron  might  be  looked  upon  as  two  square  pyramids,  base  to  base, 
the  height  of  each  being  equal  to  half  the  diagonal  of  its  base.     In  other  words, 


Fig.  68. 


the  solid  has  three  diagonals  at  right  angles  to  one  another,  and  each  one  therefore 
at  right  angles  to  the  plane  of  the  two  others. 

The  simplest  way  to  arrange  this  solid  is  shown  at  (iv)  in  Fig.  68,  where  the 
full  height  in  the  elevation  is  made  equal  to  a  diagonal  o^  the  square  representing 
the  solid  in  plan. 

At  (v)  the  solid  is  shown  cast  over  on  to  one  of  its  faces,  and  in  plan  a  dotted 
triangle  for  the  under  face,  equilateral,  and  another,  clear-lined  for  the  upper  face 
will  be  necessary.     All  the  sloping  edges  are  seen  in  plan  and  form  a  regular  he.xagon. 

In  Fig.  69  an  octahedron,  of  given  edge  ah,  is  arranged  so  that  one  diagonal 
is  in  the  V.P.  at  if  to  the  H.P.,  and  another  diagonal  is  at  45°  to  the  H.P. 


68 


DESCRIPTIVE   GEOMETRY 


From  c'  the  half  diagonal,  obtained  from  the  square  on  ah,  is  marked  off  on 
each  of  the  inclined  lines.  When  the  plans  and  elevations  of  these  have  been 
determined,  they  are  extended  to  make  the  complete  diagonals.  The  third  diagonal 
is  then  found,  as  a  perpendicular  to  the  plane  of  the  two  already  found.  Par- 
ticular care  must  be  taken  in  clear-lining  the  projections,  so  that  the  correct  use 
of  dotted  lines  is  made  for  the  edges  not  in  view.  It  should  be  recognized  that 
dd'  is  the  near  and  upper  end  of  the  third  diagonal. 


Fig.  69. 


EXERCISE   XXIX 


1.  Find  the  projections  of  a  tetrahedron,  2"  edge,  when  one  edge  is  at  25°  to  the  H.P.  and 
another  edge  is  at  45°  to  the  H.P. 

2.  Find  the  projections  of  an  octahedron,  \\"  edge,  when  one  diagonal  is  in  the  V.P.  at 
20°  to  the  H.P.,  and  another  diagonal  is  at  35°  to  the  H.P. 

3.  Find  the  projections  of  an  octahedron  \\"  edge,  when  one  edge  is  in  the  V.P.  at  20°  to 
the  H.P.  and  a  diagonal,  adjacent  to  it,  is  at  40°  to  the  H.P. 

4.  Find  the  plan  of  an  octahedron,  \\"  edge,  with  one  face  in  the  H.P.,  and  show  an  ele- 
vation of  it,  derived  from  this  plan,  on  a  plane  to  which  no  edge  of  the  solid  is  parallel. 


CHAPTER  X 

AXO.METRIC  AND  ISOMKTRIC  PROJKCTIOX 

Section  20.  Realize  experimentally  or  otherwise  the  following  facts: — 

1.  When  two  inclined  lines  meet,  there  is  a  plane  of  these  lines  which  has  its 
H.T.  passing  through  the  H.T.'s  of  the  two  lines; 

2.  When  a  line  is  perpendicular  to  a  plane,  the  plan  of  the  line  and  the  H.T. 
of  the  plane  are  at  right  angles  to  one  another; 

3.  Three  lines  may  meet  in  a  point,  and  each  one  be  perpendicular  to  both 
of  the  others,  that  is,  to  the  plane  of  the  others;   and 

4.  Three  such  lines,  each  perpendicular  to  the  others,  may  either  be  equally 
inclined  to  the  H.P.  or  all  differently  inclined. 


Fig.  70. 


Let  there  be  three  lines,  .1.  B  and  C,  meeting  in  one  pi)int.  and  containing 
right  angles  with  each  other.  If  they  be  equally  inclined  t  >  the  H.P.,  the  three 
right  angles  will  appear  in  plan  equal  to  one  another,  and  the  plans  of  the  right 
angles  will  be  1 20°  in  each  case,  as  at  (i)  in  Fig.  70. 

Now,  if  the  point  where  A,  B  and  C  meet  each  other  be  at  some  distance 
above  the  H.P.,  then  some  point  in  one  of  them,  say,  //  in  .1.  will  be  the  H.T.  of 
line  .1,  and  the  H.T.  of  B  and  the  H.T.  of  C  can  then  be  placed  similarly  in  the 
lines  B  and  C  at  the  same  distance  from  the  point  where  they  all  meet,  as  seen 
at  (ii). 


70 


DESCRIPTIVE   GEOMETRY 


By  drawing  lines  through  these  H.T.'s  we  have  the  H.T.'s  of  the  planes  of 
pairs  of  lines,  and  it  will  be  seen  that  as  the  hnes  A  and  B  contain  a  right  angle, 
they  will,  when  rabatted,  expose  that  right  angle,  which  is,  of  course,  the  angle 
of  a  semicircle;  therefore,  taking  advantage  of  this  fact,  on  HH  make  a  semicircle, 
and  by  a  perpendicular  across  the  trace  HH  from  the  plan  of  the  point  where  A 
and  B  meet,  to  the  circumference  of  the  semicircle,  we  obtain  the  rabattement  of 
A  and  B  as  seen  at  ^2  and  Bo,  and  so,  true  lengths  are  now  shown  for  these  lines 
A  and  B. 

Next  consider  the  case  when  the  lines  A,  B  and  C  are  at  different  angles  to 
the  H.P.,  and  let  A  be  at  25°  and  B  be  at  45°,  then  by  the  method  explained  and 
illustrated  in  Fig.  65,  (ii),  the  plans  of  A  and  B  are  found  and  their  H.T.'s  are 
at  HH  in  Fig.  71.     Through  HH  draw  the  H.T.  of  the  plane  of  A  and  B,  and  the 


Fig.  71 


plan  of  C  will  be  perpendicular  to  this  H.T.  Hne.  This  is  shown  by  the  arrow  line 
in  the  figure.  The  H.T.  of  this  arrow  Hne  can  now  be  found  by  drawing  the  H.T. 
of  the  plane  of  B  and  C  perpendicular  to  the  plan  A,  or  by  drawing  the  H.T.  of 
the  plane  of  A  and  C  perpendicular  to  plan  B.     This  H.T.  of  C  is  indicated  at  HC. 

From  point  HC  to  the  point  where  the  three  hnes  meet  is  the  plan  of  C,  and 
this  plan  may  be  swung  into  XY  and  then  the  elevation  of  C,  on  the  V.P.,  may  be 
drawn,  to  show  the  angle,  marked  in  the  figure  with  an  arrow-headed  arc,  just 
in  the  way  that  the  lines  A  and  B  showed  their  inclinations,  25°  and  45°  respectively, 
when  originally  placed  in  the  V.P. 

As  in  the  previous  case.  Fig.  70,  (ii),  since  each  pair  of  lines  contains  a  right 
angle,  true  lengths  for  ^,  B  and  C  may  readily  be  arrived  at  by  rabatting  them 
into  semicircles,  as  shown  at  A2,  B2  and  C2,  Fig.  72.  Upon  these  rabattements 
any  true  length  measurements  may  be  made  as  at  ^2  and  then  carried  perpendicu- 
larly across  to  the  plan  line  A  Sit  d. 


AXOMETRIC  AND   ISOMETRIC   PROJECTION 


71 


This  arrangement  of  three  lines,  A,  B  and  C,  provides  a  very  useful  means  of 
making  single  projections  or  plans  which  show  three  dimensions  to  defmite  scales, 
and  because  of  the  three  lines  which  serve  as  axes  of  direction  for  measurable 
distances,  the  method  is  known  as  Axometric  Projection. 

It  will  be  recognized  that  the  "  scale  "  of  measurement  along  each  of  the 
three  lines,  A,  B  and  C,  or  axes  of  projection,  as  they  are  called,  in  Fig.  70  at  (ii), 
is  the  same,  because  the  axes  are  equally  inclined  to  the  projection  plane.  This 
is  therefore  referred  to  as  Isometric — a  special  case  of  axometric  projection.  The 
scales  of  measurement  along  the  axes  A,  B  and  C  in  Figs.  71  and  72,  however,  are 
all  different  from  one  another,  so  that,  for  instance,  2"  on  the  axis  A  is  represented 
by  a  longer  line  than  2"  on  the  axis  B. 

As  an  illustration  of  the  use  of  this  method,  let  it  be  required  to  find  the  axo- 
metric projection  of  a  skeleton  cube,  each  face  of  which  will  appear  as  a  foreshort- 


FiG.  72. 


ened  view  of  Fig.  73,  (,i).  There  will  be  twelve  bars  of  square  section,  each  one 
represented  by  three  lines,  as  suggested  at  (ii).  There  are  three  directions  for  the 
edges  of  a  cube,  four  edges  to  each  direction,  therefore  the  three  axes  of  projection 
may  each  contain  an  edge.  Let  two  of  the  axes  be  at  25°  and  45°  respectively, 
to  the  projection  plane. 

First  find  the  axes  .1,  B  and  C  and  rabattements  of  them  at  .!_',  B2  and  €2- 
On  these  rabattements  mark  oft"  measurements  taken  from  the  edges  of  the  square 
at  (i),  and  transfer  them  to  the  axes  by  perpendiculars  to  the  trace  lines.  Then 
by  parallels  the  three  upper  faces  of  the  cube,  all  foreshortened  in  appearance, 
may  be  found.  Three  of  the  bars  are  by  this  time  represented,  each  by  three  lines. 
By  the  careful  use  of  parallels  each  of  the  remaining  nine  bars  may  be  found.  All 
the  twelve  bars  have  square  ends  which  may  be  shown  as  at  fii)  in  the  figure. 


72 


DESCRIPTIVE   GEOMETRY 


The  drawing  here  shown  is  not  completed,  but  should  be  carried  out  on  a  large 
scale  and  completed  by  the  student. 


(II) 


Fig.  73. 


EXERCISE  XXX 


1.  Find  the  projection  of  a  2"  cube  when 
two  of  the  axes  of  projection  are  inclined  at  25° 
and  40°  respectively  to  the  projection  plane. 

2.  Find  the  inclination  to  the  projection 
plane  of  the  third  axis  in  i. 

3.  Find  by  axometric  projection  a  skeleton 
cube  3I"  outside,  2\"  inside,  measurement.  One 
axis  is  at  30°  and  another  at  40°  to  the  projection 
plane. 

4.  Find  in  isometric  projection  the  two  parts, 
separated  as  in  the  illustration  herewith,  of  a 
mortise  and  tenon  joint.  Use  dotted  lines  to 
represent  those  lines  not  in  view. 


AXOMETRIC   PROJECTION,    CONTINUED 


73 


AXOMETRIC  PROJECTION,  Continued 

Section  21.  Since  "  scale,"  with  respect  to  drawings,  is  the  ratio  of  projection 
length  to  real  length,  it  will  be  seen  that  the  axes  in  axometric  projection  may  be 
arranged  in  place  when  scales  for  any  two  of  them  are  given,  by  setting  up  two 
lines  at  inclinations  to  the  H.P.  that  will  give  those  scales,  and,  having  arranged 
their  plans  correctly,  placing  the  third  axis  in  proper  relation  to  these.  The  scale 
of  the  third  may  then  be  obtained  also. 

Thus,  suppose  the  scales  for  A  and  B  are  to  be  f  and  -^  respectively,  the  incli- 
nations of  A  and  B  to  the  H.P.  may  be  obtained  as  in  Fig.  74  at  (i),  where  ab, 
equal  to  4  divisions,  is  represented  in  plan  by  ac  equal  to  3  of  the  same  divisions; 


Fig.  74. 


that  is,  ac  is  f  full  length,  and  the  inclination  for  this  scale  of  f  is  the  angle  bac. 
Again,  ad,  equal  to  6  divisions,  is  represented  in  plan  by  ac,  equal  to  5  divisions, 
i.e.,  ae  is  i  full  length  and  the  angle  for  the  axis  which  will  have  a  scale  of  ,^  will 
be  the  angle  dac. 

Employing  the  method  as  shown  in  Fig.  71,  set  up  at  Fig.  74,  (ii),  the  angle 
dae  at  a  and  the  angle  bac  at  /3,  and  proceed  to  find  axes  A  and  B  with  their  H.T.'s 
at  HH.  The  third  axes,  C,  will  be  found  and  its  H.T.  obtained  as  in  Fig.  71.  By 
swinging  the  plan  of  that  portion  of  it  above  the  H.P.  into  AT,  the  plan  length 
RS  and  the  real  length  TS  are  obtained,  which  will  give  the  scale  for  the  third  axis 

RS 


C  as 


TS' 


From  what  was  seen  previously  (see  Note  in  Section  18),  the  contained  angle 
between  the  axes  of  projection  being  90°,  the  inclinations  of  them  to  the  projection 
plane  must  be  small,  so  that  the  sum  of  them  is  less  than  00°,  and  this  limits  the 


74 


DESCRIPTIVE   GEOMETRY 


use  of  axometric  projection  to  large  scales  only,  if  the  drawings  are  to  serve  the 
purpose  they  are  intended  to  serve. 

Suppose  the  arrangement  of  the  axes  of  projection  to  be  given  in  plan  as  at 
(i),  Fig,  75,  and  let  it  be  required  to  find  the  projection  of  some  solid  having  its 
edges  or  other  lines  in  the  directions  of  the  axes  given.  First  choose  any  point 
in  one  of  the  lines,  say,  H,  in  axis  A  at  (i) ,  and  proceed  to  find  points  in  B  and  C 
that  are  at  the  same  level,  by  lines  from  H  perpendicular  to  C  and  B  respectively. 
Then  by  rabatting  the  right  angle  contained  by  axes  A  and  B,  also  by  A  and  C, 
into  semicircles,  true  lengths  are  obtained  at  ^2,  Bo  and  C2,  and  the  measurements 
of  the  cube  or  other  solid  may  be  marked  on  these  and  transferred  to  A,  B  and  C 
respectively,  as  in  Fig.  73  at  (iii). 


(TI) 


Fig.  75- 


If  the  incHnations  of  these  given  axes.  Fig.  75,  are  required  to  be  found,  it 
will  be  seen  that  a  vertical  plane  containing  one  of  them,  say,  axis  A ,  will  cut  the 
plane  of  the  others  in  the  Hne  RT,  see  (ii),  and  that  HR  is  at  right  angles  to  RT, 
therefore  the  vertical  semicircle  on  the  line  HT  for  diameter,  rabatted  as  at  HR2T, 
will  show  the  angle  the  axis  A  makes  with  the  projection  plane,  namely,  the  angle 
R2HR.  This  estabhshes  the  height  of  R  above  the  projection  plane.  Hence, 
if  the  H.T.'s  of  B  and  C  be  brought  around  into  the  line  HRT,  their  angles  will 
Hkewise  be  seen. 

A  further  illustration  is  given  in  Axometric  Projection  in  Fig.  76.  Let  it  be 
required  to  find  the  projection  of  a  pentagonal  pyramid  when  one  edge  of  the  base 
is  at  25°  to  the  projection  plane  and  the  axis  of  the  solid  is  at  40°  to  the  same  plane. 

First  find  the  three  axes  of  projection.  As  the  axis  of  the  sohd  has  to  be  in  the 
40°  direction,  the  base  must  be  placed  in  the  plane  of  the  two  others.  The  base, 
not  being  right  angular,  must  be  placed  in  a  rectangle  as  at  (ii).     As  AB  at  (ii) 


AXOMETRIC   PROJFXTIOX,   CONTIXUED 


75 


contains  an  edge  of  the  base,  the  Hne  AB  must  be  fitted  on  to  the  25"  axis  of  pro- 
jection. Therefore,  place  it  in  rabattement  at  /l2i^2and  draw  the  complete  figure 
contained  in  the  rectangle  BoAoCo,  and  proceed  to  carry  it  over  to  its  position 
in  plan.  Having  completed  the  base  in  plan,  find  its  centre  C,  and  from  that 
centre  draw  a  line  in  the  direction  of  the  40°  axis.  Find,  by  the  proper  method, 
the  projection  length  of  the  axis  at  coa^  or  c-zb-y,  according  to  its  length,  and  then 
mark  this  off  on  the  axis  line  of  the  pyramid  starting  from  c,  thus  giving  the  apex 
at  (/  or  h.  Join  the  apex  to  the  corners  of  the  base  and  fmish  as  usual  with  dotted 
lines  for  some  edges  as  the  case  may  require. 


Fig.  76. 


If  a  circular  hole  has  to  be  represented,  in  a  block,  for  instance,  in  axomctric 
projection,  points  should  be  obtained  in  the  circumference  by  using  diagonals  and 
parallels  to  the  sides  of  a  square  made  to  contain  the  circle,  the  sides  of  the  square 
being  made  to  follow  axis  directions  in  the  projection. 

EXERCISK   XXXI 


1.  The  plans  of  the  three  axes  of  projection  enclose  angles  of  iio^.  120°  and  i.^o"".  Find 
the  inclinations  of  the  three  axes  to  the  projection  plane. 

2.  The  scales  of  two  of  the  axes  for  a  projection  are  J^  and  j^.  Find  the  projection  of  the  axes 
and  represent  the  scale  of  the  third  axis. 

3.  P'ind  by  axomctric  projection  the  plan  of  a  regular  hexagonal  pyramid,  when  the  axis 
of  the  pyramid,  3"  long,  is  at  30°  to  the  projection  plane,  and  one  of  the  edges  of  the  base,  i" 
long,  is  at  45°  to  the  projection  plane. 


CHAPTER  XI 

SECTIONS  OF  SIMPLE  SOLIDS 

Section  22.  Sections  of  solids  are  made  by  passing  planes  through  them. 
By  rabatting  these  planes,  carrying  with  them  the  outline  points  of  the  section 
shape,  the  true  form  of  the  particular  section  may  be  obtained.  For  convenience, 
the  section  plane  is  usually  arranged  either  perpendicularly  to  the  H.P.  or  per- 
pendicularly to  the  V.P. 


Fig.  77 


Illustrations  of  the  process  are  shown  in  several  figures  now  to  be  considered. 

In  Fig.  77  a  cube  has  a  vertical  section  plane  RST  passing  through  it.  Hori- 
zontal edges  are  cut  by  it  at  a,  b  and  c,  while  at  d  two  of  the  inclined  edges  are 
cut.  Realize  that  three  inclined  faces  are  cut  and  one  vertical  one.  By  using 
S  as  a  centre  and  making  rabattement  of  all  the  points  in  which  the  plane  cuts 
the  edges,  the  true  form  is  obtained  on  the  V.P.  It  is  covered  with  hatching  or 
shading  in  the  figure. 

In  Fig.  78  a  square  pyramid  has  a  section  plane  RST  perpendicular  to  the 
V.P.,  passing  through  it.  The  slant  edges  of  the  pyramid  are  all  cut  in  points 
whose  elevations  are  marked  a'b'c'  and  d'.  By  using  S  as  centre  and  rabatting 
all  these  points  on  to  the  H.P.  the  true  form  is  obtained.  Notice  that  it  is  neces- 
sary to  have  the  plans  of  the  four  points.  These  are  readily  found  in  the  case  of 
a  and  of  d,  but  as  the  projections  of  the  edges  in  which  B  and  C  occur  are  so  nearly 

76 


SECTIONS  OF  SIMPLE   SOLIDS 


77 


perpendicular  to  XV  these  points  are  carried  by  level  lines  to  the  slant  edge  in 
which  a  occurs  and  from  their  plans  in  the  plan  of  this  edge  they  are  carried  by 
arcs  to  their  proper  places  in  the  plans  of  the  edges  they  belong  to,  and  thence 
by  perpendiculars  across  the  H.T.  line  to  their  rabattements.  The  true  form 
may  now  be  drawn  and  is  marked  in  the  figure  by  hatching. 

In  Fig.  79  a  right  circular  cylinder  has  a  section  plane  perpendicular  to  the 
V.P.,  passing  through  it  in  such  a  way  as  to  cut  the  top  horizontal  surface  and  cut 
also  a  considerable  amount  of  its  cur\'ed  vertical  surface.  Since  the  section  shape 
of  a  right  circular  cylinder  by  a  plane  inclined  to  its  axis  is  an  ellipse,  the  result  in 
this  case  will  be  part  of  an  ellipse,  as  the  shape  of  the  section.  Points  in  the  curved 
outline  of  the  true  form  are  obtained  by  choosing  straight  generator  lines,  as  they 
are  called,  on  the  curved  surface  of  the  cylinder,  marked  in  this  example  at  a,  b 


c,  d,  c,  etc.  Portions  of  the  elevations  of  these  generators  are  shown  in  order  to 
obtain  the  elevations  of  the  intersection  points  to  be  rabattcd.  In  the  figure 
the  shape  of  section  is  shown  covered  with  hatching. 

In  Fig.  80  a  right  circular  cone  has  a  section  plane  RST  passing  through  it, 
and  since  this  plane  cuts  all  the  generators  of  the  curved  surface  it  produces  an 
ellipse  for  the  shape  of  the  section.  This  shape  is  rabatted  on  to  the  H.P.  by 
using  5  as  centre.  Generators,  taken  in  such  places  as  will  give  a  good  distribu- 
tion of  points  in  the  curve,  are  drawn  at  a,  b,  c,  d,  e,  etc.  These  in  elevation  are 
seen  to  be  cut  by  the  section  plane,  and  the  points  in  which  they  are  cut  are  carried 
over  by  arcs  to  the  H.P.  It  is  necessary  to  locate  them  in  plan,  which  is  straight- 
forward work  except  in  the  case  of  generator  bb'.  The  point  in  this  generator 
must  be  carried  horizontally  to  the  generator  c,  and  its  plan  carried  back  to  the 
generator  b  by  an  arc  as  seen  in  the  figure. 


78 


DESCRIPTIVE   GEOMETRY 


Fig.  8 1  shows  a  right  circular  cone  with  a  section  plane  RST  passing  through 
it  so  as  to  give  the  hyperbola  as  a  section.  The  plane  for  the  hyperbola  is  parallel 
to  the  plane  of  two  generator  lines  of  the  cone,  whereas  the  plane  for  a  parabola 
has  only  one  generator  line  of  the  cone  parallel  to  it.  The  two  generators  parallel 
to  the  plane  of  the  hyperbola  in  this  case  are  shown  at  a  and  b,  and  their  Horizontal 
Traces  are  at  ////.  The  curve  of  the  hyperbola  is  shown  in  plan  at  cc  with  its  true 
form  by  rabattement  at  C2C2.     The  H.T.'s  of  this  curve  are  at  gg. 

Now,  if  the  line  cc  which  lies  on  the  curved  surface  of  the  cone  be  continually 
produced  in  the  same  plane  it  becomes  straight,  at  infinity,  on  opposite  sides 
of  the  cone,  and  since  the  plane  of  the  parallel  generator  lines  A  and  B  is  at  such 


Fig.  81. 


a  small  distance  from  the  plane  of  the  hyperbola,  the  tangents  to  the  hyperbola 
lines,  at  infinity,  and  the  generator  hnes  near  them  are  pairs  of  parallels.  These 
tangents  and  generators  on  the  surface  of  the  cone  at  infinity  will  meet  the  cir- 
cumference of  the  circular  section  of  the  cone  at  infinity,  at  right  angles,  and  the 
H.T.'s  of  the  planes  of  the  pairs  of  parallels  at  infinity  will  be  perpendiculars  to 
the  generators  whose  plans  are  a  and  b  respectively. 

Consequently,  on  the  Horizontal  Plane  of  projection,  from  points  //  perpen- 
diculars to  the  plans  a  and  b  will  pass  through  the  H.T.'s  of  the  tangents  at  infinity 
to  the  hyperbola,  which  tangent  lines  also  He  in  the  plane  of  the  hyperbola.  There- 
fore HH  are  the  H.T.'s  of  the  tangents  at  infinity  to  the  hyperbola,  and  the  plans 
of  them  are  the  lines  dd,  made  parallel  in  plan  to  the  generators  a  and  b. 


TRACES  OF  CURVED  SURFACES  79 

The  tangents  at  infinity  to  the  h\perbola  are  called  the  Asymptotes,  and  meet 
each  other  at  an  angle.  In  Fig.  8i  they  are  rabatted  on  to  the  H.P.  and  show  the 
angle  contained,  which  is  marked  by  an  arrow-headed  arc. 

Strictly  speaking,  the  asymptotes,  lying  outside  the  two  branches  of  a  hyper- 
bola, which  are  sections  of  two  equal  cones  united  at  their  ape.xes  and  having  the 
same  axis  line,  are  lines  which  never  really  meet  the  curves  of  the  hv-perbola. 

exp:rcise  XXXII 

1.  A  square  prism,  15"  edge  of  end,  2§"  long,  has  a  long  edge  in  the  H.P,  at  30°  to  XV, 
and  the  face  adjacent  to  it  is  at  30°  to  the  H.P.  Find  the  true  form  of  section  by  a  vertical  plane 
at  45°  to  the  \'.P.,  passing  through  the  soh'd  and  cutting  its  axis  i"  from  one  end.  Two 
solutions. 

2.  Find  the  true  form  of  section  of  a  right  hexagonal  pyramid,  base  in  H.P.  with  one  diag- 
onal of  base  parallel  to  A'F,  by  a  plane  perpendicular  to  V'.P.,  at  45°  to  the  H.P.,  and  passing 
through  a  point  in  the  axis  |"  from  the  base.     Base  edge  i".     Axis  22". 

3.  Find  the  true  form  of  a  parabola  by  a  plane  passing  through  a  cone,  when  its  base  is  2" 
diameter,  and  height  of  apex  of  the  cone  above  the  base  25".  The  plane  cuts  the  axis  in  the 
centre. 

4.  Find  the  asymptotes  and  show  the  angle  contained  by  them  in  the  case  of  a  hyperbola 
made  by  a  plane  at  15°  tp  the  axis  of  a  cone,  cutting  the  axis  in  a  point  i"  from  the  apex.  The 
angle  of  the  cone  at  the  apex  is  40°. 


TRACES  OF  CURVED  .SURFACES 

Section  23.  When  a  right  circular  cylinder  is  inclined  to  a  projection  plane, 
its  curved  surface,  produced  to  meet  the  plane,  will  meet  it  in  the  curve  of  an 
ellipse,  for  the  projection  plane  might  be  considered  as  a  section  plane  passing 
through  the  circular  cylinder  at  an  angle  to  its  axis.  This  applies  also  to  the 
curved  surface  of  a  cone.  The  line  in  which  the  curved  surface  of  the  cone  meets 
the  projection  plane  will  be  an  ellipse,  if  the  cone  is  a  right  circular  one,  i.e.,  one 
that  has  a  circular  section  in  any  plane  at  right  angles  to  its  a.xis. 

The  line  of  intersection  in  which  the  curved  surface  of  a  cylinder  or  of  a  cone, 
whether  its  axis  be  inclined  or  not.  will  meet  the  projection  plane,  if  the  curved 
surface  is  produced  to  do  so,  is  called  the  trace  of  the  curved  surface,  and  may  be 
the  Vertical  trace  or  the  Horizontal  trace  according  to  whether  it  is  found  on  the 
V.P.  or  on  the  H.P.     Illustrations  are  shown  in  Figs.  82.  and  83. 

It  will  be  seen  that  in  the  case  of  the  right  circular  cylinder,  the  H.T.  is  an 
ellipse  with  its  minor  axis  equal  to  the  diameter  of  the  cylinder,  whereas  in  the 
case  of  the  right  circular  cone  the  ellipse  for  H.T.  has  a  minor  axis  which  varies 
in  length  with  the  inclination  of  the  cone  or  with  the  distance  of  its  apex  from  the 
projection  plane.  The  method  of  obtaining  the  trace  is  to  choose  any  number  of 
generator  lines  and  produce  them  until  their  traces  are  found,  then  to  draw,  by 
freehand,  the  curve  through  these  trace  points. 


80 


DESCRIPTIVE   GEOMETRY 


A  further  illustration  is  shown  in  Fig.  84,  where  it  is  important  to  realize 
that  the  generators  whose  plans  are  at  aa  have  their  elevations  at  a'a',  and  that 


Fig.  82. 


Fig.  S3. 


these  are  best  located  by  recognizing  the  fact  that  they  pass  through  the  ends 
of  the  horizontal  diameter  of  the  end  of  the  cylinder.     The  topmost  generator 


Fig.  84. 


line,  having  its  plan  at  h,  has  its  elevation  at  //,  which  is  not  the  topmost  line  of 
the  elevation.     Other  generators,  chosen  at  convenience  for  the  purpose  of  secur- 


TRACES  OF  CURVED  SURFACES  81 

ing  points  in  the  trace,  such  as  those  marked  cc-z  in  the  figure,  should  meet  the  end 
of  the  cyUnder  in  pairs,  at  the  ends  of  horizontal  chords,  for  then  the  elevation 
marked  c'2  can  the  more  certainly  be  located  by  making  use  of  the  horizontal 
chord  through  the  end  of  the  one  marked  cc' . 

N.B. — The  projections  for  the  circular  ends  of  the  solids  should  be  carefully 
found  by  the  method  previously  discussed.     (See  Section  4,  Fig.  8.) 


EXERCISE  XXXIII 

1.  Find  the  H.T.  of  the  curved  surface  of  a  right  circular  cylinder,  i\"  diameter,  when  its 
axis  is  at  35°  to  the  H.P.  and  the  plan  of  it  is  at  25'  to  XY . 

2.  Find  the  H.T.  of  the  curved  surface  of  a  right  circular  cone,  whose  base  is  i\"  diameter 
and  axis  ih"  long.     Let  the  axis  be  parallel  to  the  V.P.,  and  inclined  to  the  H.P.  at  45°. 

3.  Show  the  plan  and  elevation  of  a  cone  whose  H.T.  is  a  circle  2^"  diameter,  and  whose 
axis  is  3"  long  and  inclined  at  40°  to  the  H.P.  and  40°  to  the  V.P.  Show  the  V.T.  of  its  curved 
surface  when  the  centre  of  the  circular  H.T.  is  i\"  from  XY. 

4.  Find  the  shadow  of  a  sphere  of  2"  diameter,  having  its  centre  2"  above  the  H.P.  and  2" 
in  front  of  the  \'.P.,  when  the  rays  of  parallel  light  are  inclined  at  45°  to  the  H.P.  and  30°  to 
the  V.P. 

N.B. — The  shadow  problem  here  given  involves  the  method  discussed  in  this  section. 


PART  THREE 


CHAPTER   XII 

-  TANGENT  PLANES  TO  CONES  AND  CYLINDERS 

Section  24.  Since  the  generator  of  a  cylinder,   and  also  of  a  cone,   is  a 
straight  line,  it  is  evident  that  a  plane,  tangent  to  or  touching  the  curved  surface 


Fig.  85. 


of  one  or  of  the  other,  will  touch  it  along  a  straight  hne,  and,  in  the  case  of  the 
cone,  will  include  its  apex.  Such  a  tangent  plane  may  be  shown,  as  usually  planes 
are  shown,  by  its  traces,  and  its  traces  will  be  tangential  to  the  curved  traces  of 
the  curved  surface  of  the  solid. 

In  the  illustration,  Fig.  85,  a  right    circular    cone,  that  is,  one  whose  sec- 
tion perpendicular  to  its  axis  is  circular,  stands  vertically  on  the  H.P.,  conse- 

82 


TANGENT   PLANES   TO   CONES   AND   CYLINDERS 


83 


quently,  the  circle,  which  is  the  circumference  of  its  base,  is  the  H.T.  of  its  cur\^ed 
surface.  A  plane,  tangent  to  this  cone,  will  therefore  have  its  H  T.  tangent  to 
this  circular  H.T..  and  this  is  shown  at  ST.  As  the  apex  aa'  is  in  the  tangent 
plane,  a  horizontal  line  may  be  drawn  through  it,  lying  in  the  tangent  plane. 
This  will  have  its  plan  parallel  to  ST,  and  its  elevation  will  give  its  V.T.  at  V. 
RS,  the  V.T.  of  the  plane,  may  then  be  drawn. 

It  will  be  realized  next,  that,  as  all  the  generators  of  the  vertical  cone  meet 
the  H.P.  at  the  same  incUnation,  therefore  the  generator  whose  plan  is  ab.  is  at 
the  same  inclination  as  the  generator  whose  elevation  is  a'c',  and  whose  inclination 
is  indicated  by  an  arrow-headed  arc  in  the  figure;  and,  since  the  plane  is  at  the 
same  inclination  as  a  line  in  it  perpendicular  to  its  trace,  and  AB  is  perpendicular 
to  the  trace  ST.  the  tangent  plane  RST  is  at  the  angle  indicated  at  c'. 


Fig.  86. 


The  converse  of  this  problem  is  specially  useful.  For  instance,  if  it  be 
required  to  fmd  the  traces  of  a  plane  of  given  inclination  to  the  H.P.  having  its 
H.T.  in  a  given  direction,  it  is  only  necessary  to  set  up  a  right  circular  cone  whose 
generators  are  at  the  given  inclination  for  the  required  plane,  and  then  make  the 
H.T.  of  the  plane  tangent  to  the  circular  trace,  the  base  of  the  cone  is  the  H.P.. 
and  proceed  as  before  to  find  the  V.T.  of  the  plane. 

In  Fig.  86  the  V.T.  of  a  plane  is  given  at  RS.  Suppose  the  plane,  whose  V.T. 
is  RS,  to  be  inclined  at  6o°  to  the  H.P.,,  and  let  it  be  required  to  find  its  H.T. 
Then,  because  any  point  in  this  V.T.  is  in  the  plane  and  may  be  taken  as  the 
apex  of  a  right  circular  vertical  cone  to  which  the  plane  is  tangent,  from  any  such 
point  A  draw  the  line  a'c'  in  the  V.P.  at  6o°  to  the  H.P.  The  circumference 
of  the  circle  drawn,  with  a  as  centre  and  ac'  as  radius,  is  the  H.T.  of  the  6o°  cone, 
and  the  point  A,  its  apex,  is  contained  in  every  tangent  plane  to  the  cone.  There- 
fore from  the  point,  namely  5.  in  .VI',  where  the  V.T.  line  meets  it,  draw  the  H.T. 


84 


DESCRIPTIVE  GEOMETRY 


line,  ^ST,  of  the  required  60'  plane,  making  it  tangent  to  the  circular  trace  of  the 
cone.  The  line  SN  shows  the  H.T.  of  another  plane,  having  the  same  V.T., 
and  tangent  to  the  same  60°  cone,  therefore  also  at  60°  to  the  H.P.  There  are, 
therefore,  two  planes,  RST  and  RSN,  satisfying  the  conditions. 

In  Fig.  87  a  line  AB  is  given,  and  it  is  shown  how  a  plane  of  any  particular 
inclination  to  the  H.P.,  greater  than  the  inclination  of  the  given  line  to  the  H.P., 
may  be  found,  when  it  has  to  contain  the  given  Kne.  For  the  solution,  any  point 
cc'  in  the  given  line  may  be  taken,  to  serve  as  the  apex  of  a  right  circular  vertical 
cone. 


Fig.  87. 


The  cone  may  be  represented  by  drawing  the  line  c'd'  to  make  with  XY 
an  angle,  a,  equal  to  the  given  inclination  of  the  required  plane,  and  with 
cd,  its  plan,  as  radius,  drawing  the  circular  H.T.  of  its  curved  surface.  Next, 
find  the  H.T.  of  the  given  line  AB  a.t  H,  and  through  H  draw  tangent  lines  to 
the  circle.  These  are  the  horizontal  traces  of  two  tangent  planes  to  the  cone, 
each  containing  the  given  line.  Horizontal  Hnes  on  these  planes,  drawn  through 
the  apex  C,  will  give  points  V  and  Vi  in  the  Vertical  Traces  required  to  more  com- 
pletely determine  the  planes.  HeV  is  one  of  the  planes.  The  other  plane,  whose 
H.T.  is  Hf,  gives  an  opportunity  of  showing  how  to  obtain  the  V.T.  without  first 
making  Hf  meet  XY.  Thus,  from  any  point  g,  in  Hf,  draw  the  plan  of  a  line  through 
some  point  in  the  line  ^^— through  the  apex  C  will  do— and  produce  this  plan 


TAXGEXT  PLANES  TO  COXES  AXD  CYLIXDERS 


85 


and  its  elevation  till  its  V.T.  is  found  at  V2.     Join  V1V2  and  this  line  is  part  of 
the  Ventcal  Trace  of  the  second  plane,  part  of  whose  H.T.  is  the  line  II f. 


EXERCISE   XXXIV 

1.  Find  a  plane  at  50°  to  the  H.P.  containing  ihe  given  point  .1  at  (i).     Let  the  H.T.  of  the 
plane  be  at  45°  to  the  A' I'. 

2.  Find  a  plane  at  65^  to  the  H.P.  and  containing  the  given  line  AB  at  (iij. 

3.  The  \'.T.  of  a  plane  is  given  at  (iii).     Show  the  H.T.  of  it  when  the  plane  is  at  45''  to  the 
H.P.     Two  solutions. 


(i 

(ii)    /; 

(iii) 

.  Ta' 

/ 

4 

d( 

1" 

^^'^ 

X 

' 

:!(i^\ 

Y 

-'' 

a\ 

\»/ 

'  ia 

\ 

^ 

By  experiment  with  a  cylindrical  object — a  roll  of  paper  will  do — it  should 
be  realized  that  if  the  cylinder  is  inclined,  there  may  be  any  number  of  planes 
tangent  to  it,  of  angles  to  the  H.P.  not  less  than  the  angle  of  inclination  of  the 
axis  of  the  cylinder  to  the  H.P. 

Thus,  in  Fig.  88  at  (i)  the  plane  RST  is  tangent  to  the  inclined  cylinder  and 
is  at  the  same  inclination  to  the  H.P.  as  that  of  the  cylinder.  Any  plane  of  greater 
inclination  may  be  made  tangent  to  the  same  cylinder,  until  one  is  reached  that 
is  vertical,  such  as  that  whose  H.T.  is  at  LL.  If  the  cylinder  were  inclined  to  the 
V.P.  also,  then  this  plane  LL  would  have  a  V.T.  perpendicular  to  the  A' I'. 

As  previously  noted,  the  condition  for  a  plane  to  be  tangent  to  a  cylinder 
is,  that  it  shall  contain  a  straight-line  generator  of  the  curved  surface,  and  as  all 
such  generators  have  their  H.T.'s  in  the  cur\'e  of  the  ellipse  serving  for  the  H.T. 
of  the  cylinder,  the  H.T.  of  any  tangent  plane  to  the  cylinder  must  be  a  tangent 
line  to  the  curved  H.T.  of  the  surface  of  the  cylinder. 

Now  let  ah.  a'h\  at  (ii),  be  the  projections  of  a  line  parallel  to  the  generators 
of  the  given  cylinder,  and  from  some  point  in  it,  serving  as  the  apex  of  a  cone, 
arrange  a  right  circular  vertical  cone  whose  generators  are  at,  say  a°  to  the  H.P. 
Two  a°  tangent  planes  may  now  be  represented  by  their  H.T.'s  at  ac  and  ad, 
to  contain  the  line  AB;  and  because  parallel  planes  have  their  H.T.'s  parallel, 
and  AB  is  a  parallel  to  all  the  tangent  planes  to  the  given  cylinder,  being  parallel 


86 


DESCRIPTIVE   GEOMETRY 


to  all  its  generators,  there  will  be  four  a°  planes,  two  parallel  to  plane  ac  and  two 
parallel  to  ad,  and  they  will  be  shown  by  their  H.T.'s,  tangent  to  the  curved  H.T. 


Fig.  88. 


of    the  cylinder,  as  at  e,  f,  g  and  h.     Each  of  these  four  planes  contains  one 
generator  of  the  cylinder,  and  therefore,  since  the  generators  of  the  cylinder  are 


Fig.  89. 


parallel  to  the  V.P.,  the  V.T.'s  of  these  planes  will  be  parallel  to  the  elevations  of 
the  generators  of   the  cylinder.     Therefore,   by  producing   the  H.T.   marked  g 


TAXGEXT   PLANES  TO   CONES   AND   CYLINDERS 


87 


to  XY,  the  V.T.  for  it  may  be  drawn  as  shown,  parallel  to  the  generators  of  the 
cylinder.  So,  also,  with  the  V.T.'s  of  the  other  planes.  X.B.  The  ellipse  for  H.T. 
should  be  obtained  by  the  method  shown  in  Section  23. 

In  Fig.  89  is  shown  a  right  circular  cylinder  whose  axis  is  inclined  to  both 
planes  of  projection.     Its  H.T.,  an  ellipse,  is  also  shown.     Suppose  a  tangent  plane 


Fig.  90. 


to  this  cylinder  has  ST  for  its  H.T.,  obtained  as  in  the  previous  case;  then  because 
the  point  H  is  the  H.T.  of  a  generator  of  the  cylinder,  and  //'  is  the  elevation  of 
//,  the  generator  having  //  for  its  H.T.  can  be  located,  as  shown  in  the  figure  at 
GG',  and  produced  until  its  V.T.  is  found  at  V.  Now  draw  VSR,  which  is  the 
V.T.  of  the  tangent  plane.   So  proceed  with  others. 

In  Fig.  90,  a  right  circular  cone  is  represented  in  plan  and  elevation,  with 
axis  inclined  to  both  planes  of  projection.     Its  H.T.,  the  large  ellipse,  is  also  shown. 


88  DESCRIPTIVE   GEOMETRY 

Let  it  be  required  to  find  the  traces  of  planes  tangent  to  this  inclined  cone  and 
having  a  given  inclination  to  the  H.P.  Since  all  planes,  tangent  to  the  given  cone, 
must  pass  through  the  apex  of  it,  and  a  plane  of  a  particular  inclination,  say  a°, 
must  be  tangent  to  a  vertical  right  circular  cone  whose  generators  are  at  that 
particular  inclination,  it  is  necessary  to  make  use  of  the  apex  of  the  given  inclined 
cone  to  serve  as  the  apex  of  such  a  vertical  one.  The  H.T.  of  the  vertical  cone 
referred  to  is  the  circle  H  in   the  figure. 

It  should  now  be  realized  that  a  plane  tangent  to  both  cones  at  the  same  time, 
will  have  its  H.T.  as  a  common  tangent  to  the  H.T.'s  of  the  two  cones,  and  will 
contain  the  common  apex  A.  The  H.T.  of  such  a  plane  is  shown  at  ST,  and  to 
obtain  the  point  V  in  its  V.T.,  a  horizontal  line,  on  the  tangent  plane,  may  be 
drawn  through  the  apex  aa'.     The  line  SV  is  the  V.T.  required. 

In  the  case  illustrated  in  Fig.  90,  three  other  tangent  planes  of  the  same 
inclination  may  be  found.  If  the  H.T.  of  one  of  them  is  parallel,  or  nearly  parallel, 
to  the  XY,  or  at  such  a  small  angle  with  it  that  there  is  not  room  for  the  point 
to  be  located  in  XY,  where  the  H.T.  and  the  V.T.  meet,  then  two  inclined  Hues 
on  the  plane  required  may  be  drawn  through  the  apex  and  through  any  convenient 
points  c  and  d  in  the  H.T.,  and  their  V.T.'s  found  at  Vi  and  V2.  The  line  drawn 
through  these  V.T.  points  is  the  V.T.  of  the  tangent  plane  whose  H.T.  is  the 
line  upon  which  the  points  c  and  d  were  chosen. 


EXERCISE   XXXV 

1.  Find  the  four  tangent  planes,  each  at  65°  to  the  H.P.,  to  a  right  circular  cylinder  whose 
axis  is  at  40°  to  the  H.P.  and  at  25°  to  the  V.P.  Diameter  of  cylinder  i|".  Let  the  cylinder 
rest  on  a  point  in  the  H.P.  2"  from  XY. 

2.  Find  the  traces  of  planes  inclined  to  the  H.P.  at  65°  and  tangent  to  a  right  circular  cone 
of  iV'  base,  and  ih"  axis,  when  the  axis  is  at  40°  to  the  H.P.  and  the  plan  of  the  axis  is  at  ■ 
45°  to  XY.    Let  the  cone  rest  on  the  H.P.  in  a  point  15"  from  the  XY. 


PROJECTION  OF  SOLIDS  DEPENDENT  ON  TANGENT  PLANES  TO 
RIGHT  CIRCULAR  CONES 

Section  25.  In  finding  the  projections  of  prisms,  cubes  and  pyramids  when 
different  faces  of  the  same  soHd  are  at  different  inclinations  to  the  same  projec- 
tion plane,  it  is  necessary  to  make  use  of  the  plane  tangent  to  a  right  circular  cone. 
Illustrations  are  given  and  explained  in  Figs.  91  and  92. 

In  Fig.  91  the  method  is  shown  for  finding  the  plan  of  a  cube,  or  of  a  right 
prism,  when  one  face  is  at,  say,  40°  to  the  H.P.  and  another  face  is  at,  say,  70° 
to  the  H.P. 

RST  is  the  40°  plane  in  which  one  of  the  faces  will  be  found,  and  is  arranged 
perpendicularly  to  the  V.P.  for  greater  convenience.     The  line  AB,  shown  in 


PROJECTION  OF  SOLIDS  DEPENDENT  ON  TANGENT  PLANES    89 

plan  at  ab  and  in  elevation  at  a'b',  is  an  edge  of  the  cube  perpendicular  to  the  40 "" 
face  and  therefore  may  serve  as  an  edge  of  the  70°  face. 

The  70°  plane,  to  include  this  edge  of  the  70°  face,  is  found,  according  to 
the  method  recently  explained,  by  the  use  of  a  cone  of  70°  with  its  apex  in  the  line 
AB  or  that  line  produced.  This  70°  plane,  thus  containing  AB,  has  its  H.T. 
at  ////. 


Fig.  91. 


The  point  A  is  common  to  both  planes,  the  40°  and  the  70°;  so,  also,  is  the 
point  //  where  the  two  planes'  H.T.'s  meet.  Join  //  to  a  therefore,  and  so  obtain 
the  plan  of  the  intersection  of  the  40°  plane  with  the  70°  plane.  This  line  will  con- 
tain the  edge  of  the  cube  common  to  the  two  faces  whose  inclinations  are  given. 
Rabatte//a  to  ha2,  and  using  ao  as  the  comer  for  a  square,  mark  off  J2C2  on  the 
rabatted  line,  and  also  make  d-ido  perpendicular  to  it.     These  measurements,  in 


90 


DESCRIPTIVE   GEOMETRY 


the  case  of  a  cube,  will  be  the  same  as  that  of  a'b'  previously  chosen  as  the  length 
of  an  edge  of  the  cube,  or  they  should  be  made  the  same  as  the  edges  of  the 
40°  face  of  the  prism,  if  it  be  a  prism  that  has  to  be  projected. 


Fig.  02. 


It  will  readily  be  seen  that  from  this  rabattement  of  the  two  edges  the  plans 
ac  and  ad  may  be  obtained.     Three  edges  are  now  found  in  plan,  namely  AB^ 


PROJECTION  OF  SOLIDS  DEPEXDEXT  OX  TAXGEXT  PLAXES    91 

AC  and  AD,  and  parallels  to  these  will  be  necessary  in  order  to  complete  the 
plan  of  the  cube. 

Of  the  two  faces  found  in  the  figure,  the  face  dace  is  a  40°  face,  and  the  face 
bacf  is  a  70°  face. 

In  Fig,  92  is  shown  the  case  of  a  pyramid  in  which  the  mjth  ;d  requires 
the  employment  of  two  right  circular  cones,  since  the  triangular  faces  of  the  solid 
are  inclined  to  the  plane  of  the  base. 

Let  it  be  required  to  find  the  plan  of  a  square  pyramid  when  the  base  is  inclined 
at,  say,  45°  to  the  H.P.  and  one  triangular  face  is  at,  say,  65°  to  the  H.P.  The  base 
plane  is  marked  RST  in  the  figure,  and  is  purposely  arranged  perpendicularly 
to  the  V.P.  in  order  to  simplify  the  work.  At  (i)  in  the  figure  the  square  pyramid 
is  set  up  in  plan  and  elevation  in  such  a  way  as  to  find  the  angle,  a'^,  of  the  vertical 
cone  to  which  each  of  the  triangular  faces  of  the  solid  is  tangent.  Each  face  is 
tangent  on  a  generator  which  is  at  the  same  time  the  middle  line  of  a  triangular 
face  and  meets  the  middle  point  of  a  base  edge  of  the  pyramid,  as  at  d  in  be.  At 
any  convenient  place  on  the  plane  i^^T  at  (ii)  set  up  the  right  cone  with  base  angle 
equal  to  that  in  (i),  namely  a°.  Then  find  its  horizontal  trace,  the  large  ellipse, 
on  the  H.P.  A  tangent  plane  to  this  inclined  cone,  with  its  inclination  accord- 
ing to  that  given  in  the  problem  for  a  triangular  face  of  the  pyramid,  viz.,  65°, 
is  now  obtained,  with  its  H.T.  at  HH2.  This  trace  is  made  tangent  to  the  large 
ellipse  and  at  the  same  time  tangent  to  the  circular  trace  of  a  vertical  cone  of  65° 
from  the  same  apex  as  that  of  the  inclined  cone,  viz.,  the  point  aa'.  The  V.T. 
of  this  plane  is  not  needed,  so  is  therefore  neglected. 

The  generator  ae,  a'e\  of  the  inclined  cone,  which  at  the  same  time  lies  in  the 
65°  plane  is  now  drawn,  e  being  its  H.T.  Its  elevation  ae  crosses  RS,  giving  the 
elevation  d' ,  of  the  point  in  which  it  passes  through  the  base  plane.  The  line  ad  in 
(ii)  corresponds  to  the  ad  in  (i),  and  by  drawing  a  line  through  d  and  the  point  // 
where  the  H.T.'s  of  the  45°  and  65°  planes  intersect,  the  plan  of  the  intersection 
of  the  two  planes,  base  plane  and  face  plane,  is  obtained.  The  rabattement  of  thij 
intersection  line  is  ///.  On  this  line  /7/mark  off,  on  either  side  of  the  rabattement 
of  point  d,  marked  d-i,  the  measurements  of  dh,  dc  from  the  base  edge  in  (i),  and 
construct  the  square  in  rabattement.  Bring  this  up  to  its  proper  place  in  plan, 
where  it  will  appear  foreshortened,  and  the  edges  of  it  will  be  tangents  to  the  small 
ellipse,  the  plan  of  the  base  of  the  inclined  cone.  Join  the  corners  of  the  square 
to  the  apex  a,  and  the  plan  will  be  completed  when  such  edges  as  are  not  in  view 
are  shown  by  dotted  lines.  In  the  figure  the  plan  of  the  solid  is  not  completed, 
in  order  to  save  confusion  of  lines. 

N.B. — In  the  figure  the  construction  work  for  finding  the  ellipses  is  reduced 
or  omitted  as  far  as  possible  to  save  confusion  of  lines.  In  finding  the  plan  of  the 
base  of  the  inclined  cone,  however,  there  is  shown  a  method  of  finding  plans  of 
horizontal  chords  of  it.  Point  g  is  the  elevation  of  such  a  chord.  Its  plan  can 
be  found  by  making  a  semicifcle  as  shown,  and  a  perpendicular  from  g  in  the 
diameter  of  it.  to  the  circumference,  thus  giving  half  the  chord  length. 


92  '  DESCRIPTIVE   GEOMETRY 


EXERCISE   XXXVI 

1.  Work  out  completely  the  problem  in  Fig.  gi,  and  obtain  an  elevation  of  the  solid  on  a 
vertical  projection  plane  whose  XY  is  at  45°  to  the  XY  given.  Let  the  length  of  the  edge  of  the 
cube  be  2". 

2.  Work  out  completely  the  problem  in  Fig.  92,  making  the  base  edge  of  the  pyramid  2" 
and  its  height  3I",  with  the  triangular  face  at  70°  to  the  H.P.  instead  of  at  65°. 

3.  Find  the  plan  and  elevation  of  a  square  prism,  when  the  square  end  of  it  is  at  45°  to  the 
H.P.  and  a  rectangular  face  is  at  70°  to  the  H.P.     Short  edges  il",  long  edges  2^". 

4.  Find  the  plan  of  a  regular  pentagonal  pyramid,  i"  edge  of  base,  height  3",  when  the  base 
is  incHned  to  the  H.P.  at  40°,  and  one  triangular  face  is  at  65°  to  the  H.P. 


CHAPTER   XIII 

TANGENT  PLANES  TO  A  SPHERE,  AND  THE  FINDING  OF  AN  OBLIQUE  PLANE 
WITH  GIVEN  INCLINATIONS 

Section  26.  In  discussing  tangent  planes  to  spheres  it  must  chiefly  be  noticed 
that  the  radius  of  the  sphere,  from  the  tangent  point  on  its  surface,  is  a  perpen- 
dicular to  the  tangent  plane,  and  its  projections  will  therefore  be  perpendicular  to 
the  traces  of  the  tangent  plane. 


yiC:  93- 


ru-..  94. 


In  the  illustration.  Fig.  93,  a  plane  is  shown  tangent  to  a  given  sphere  the 
centre  of  which  is  aa'.  The  radius,  which  is  perpendicular  to  the  plane,  is  also 
shown  in  elevation  and  plan,  a'p'  and  up.  A  great  circle  of  the  sphere,  that  is, 
one  whose  plane  passes  through  the  centre  of  the  sphere,  is  shown  in  plan  by  the 
(lotted  line  be  and  in  elevation  by  the  circle  b'c'.  It  is  perpendicular  to  the  tan- 
gent plane,  and  has  for  its  radius  the  line  AP. 

In  Fig.  94,  any  point  on  the  surface  of  the  sphere  is  shown  in  plan  at  p.  It 
is  required  to  find  the  tangent  plane  to  the  sphere  at  this  point.  Any  number 
of  points  on  the  surface  of  the  sphere,  and  at  the  same  level,  would  be  in  the  cir- 
cumference of  a  horizontal  circle  whose  eU^vation  would  appear  as  a  straight  line 

93 


94 


DESCRIPTIVE   GEOMETRY 


a'b'.     Consequently,  such  a  horizontal  circle  is  represented  and  p'  found  in  its 
elevation. 

Similarly,  if  the  point  be  thought  of  as  being  on  the  circumference  of  a  vertical 
circle  whose  plan  would  be  a  straight  line  through  p  parallel  to  XY,  a  circle  in 
elevation  would  need  to  be  drawn  in  order  to  locate  the  elevation  of  the  point. 
A  tangent  plane  to  the  sphere  at  P  may  now  be  obtained  by  drawing  a  horizontal 
line  through  P  in  the  direction  of  the  H.T.  required,  i.e.,  perpendicular  in  plan, 
to  the  plan  of  the  radius  from  P.     Thus  will  be  obtained  a  point  V  in  the  V.T. 


Fig.  95. 


required,  and  the   traces  of   the  plane  may  now  be  drawn  perpendicular  to  the 
radius  from  P. 

In  Fig.  95  the  same  problem  as  in  Fig.  94  is  solved  by  the  employment  of  a 
vertical  plane  containing  the  radius  from  the  given  point  and  therefore  at  right 
angles  to  the  plane  required.  This  vertical  plane  has  HH  for  its  H.T.  By  rabatting 
it  on  to  the  H.P.,  the  centre  of  the  sphere  is  carried  over  to  C2,  and  a  rabatted  great 
circle  of  the  sphere  is  made  on  the  H.P.,  shown  in  the  figure  covered  with  hatching. 
The  point  P  will  now  appear  rabatted  to  p2,  and  the  rabattement  of  the  radius  is 
seen  at  c-ypi-  At  right  angles  to  this  draw  poh  as  the  rabattement  of  the  intersec- 
tion of  the  required  plane  with  the  vertical  plane  employed  whose  H.T.  is  the 
line  HH.  This  gives  /?,  a  point  in  the  H.T.  of  the  required  plane.  Through  h 
draw  the  H.T.  marked  ST,  and  through  5  draw  RS  at  right  angles  to  the  elevation 


TANGENT  PLANES  TO  A  SPHERE 


95 


of  the  radius.  RST  is  the  plane  required.  The  height  of  /»'  above  XY  is  taken 
from  the  rabattement,  as  indicated  by  the  bracket  line.  This  construction,  Fig. 
95,  will  be  necessary,  used  conversely,  when  the  trace  of  a  plane  tangent  to  a  sphere 
is  given,  and  it  is  required  to  find  the  other  trace  of  the  plane  and  the  projections 
of  the  tangent  point. 

EXKRCISK   XXXVII 

1.  Find  the  planes  tangent  to  the  sphere  whose  projections  are  at  .1,  each  containing  a 
point  on  its  surface,  for  which  />'  is  the  elevation. 

2.  A  sphere,  2"  diameter,  touches  both  planes  of  projection.     Find  the  two  planes,  each 
touching  the  sphere  in  a  point  15"  above  the  H.P.  and  i§"  from  the  V.P. 

3.  A  sphere  is  given  at  B,  and  the  H.T.  of  a  plane  tangent  to  it.     Find  the  V'.T.  and  also 
show  the  plan  and  elevation  of  the  tangent  [)oint. 


4.  Find  the  traces  of  a  plane  tangent  to  the  sphere  whose  projections  are  at  C,  and  make 
it  perpendicular  to  the  line  AB,  whose  projections  are  given.  Mark  the  plan  and  elevation 
of  the  tangent  point. 

5.  Find  the  traces  of  three  planes  equally  inclined  to  each  other  and  all  at  60^  to  the  H.T. 
Let  ihcm  be  tangent  to  a  sphere  of  2"  diameter,  resting  on  the  H.P.  with  centre  i^"  from  the  \'.P. 
Find  also  the  inclination  between  any  two  of  the  tangent  planes. 


By  e.xperiment  and  examination  it  should  be  realized  that  a  sphere  may  be 
enveloped  by  a  right  circular  cone,  having  for  its  apex  any  point  outside  the 
sphere,  and  that  if  two  cones  envelope  the  same  sphere,  a  tangent  plane  to  the 
sphere  may  contain  both  apexes;  also,  that  any  two  unequal  spheres  may  be  en- 
veloped by  a  cone,  the  apex  of  which  will  lie  in  the  line  passing  through  their  centres. 
The  tangent  lines  on  the  surfaces  of  such  spheres  as  are  enveloped  by  cones  are 
circumferences  of  circles,  smaller  than  great  circles  of  the  spheres,  and  perpendicular 
to  the  axis  of  the  cone  in  each  case. 


96 


DESCRIPTIVE  GEOMETRY 


Fig.  96  is  an  illustration  of  the  use  of  cones  enveloping  a  single  sphere,  and 
is  a  second  method  for  obtaining  the  traces  of  an  oblique  plane  whose  angles 
with  the  planes  of  projection  are  given.  The  previously  discussed  and  more  com- 
monly used  method  was  explained  in  Part  I,  Section  9,  and  was  shown  to  depend 
on  the  projections  of  a  Hne  set  up  at  angles  complementary  to  those  of  the  required 
plane,  the  traces  of  the  required  plane  being  then  drawn  perpendicular  to  the  pro- 
jections of  the  line. 

In  the  method  now  discussed,  a  sphere  with  its  centre  in  the  XY,  is  represented 
by  the  circle  with  centre  at  c.  This  circle  therefore  serves  for  both  plan  and 
elevation  of  the  sphere. 


Fig.  96. 


Suppose  the  required  plane  is  to  have  angles  of  60°  to  the  H.P.  and  45°  to 
the  V.P.,  then  a  vertical  cone  to  envelope  the  sphere,  with  its  generators  at  60"*, 
to  the  H.P.  and  its  apex  therefore  at  V  in  the  V.P.,  will  have  part  of  its  circular 
H.T.  represented  by  the  arc  de,  and  any  plane,  whose  H.T.  is  a  tangent  to  this 
arc,  and  whose  V.T.  contains  the  apex  V,  is  therefore  a  plane  at  60°  to  the  H.P. 
Next,  represent  another  enveloping  cone,  about  the  same  sphere,  with  its  axis 
perpendicular  to  the  V.P.,  and  its  apex  therefore  at  the  point  H  in  the  H.P.,  and 
make  its  generators  at  45°  to  the  V.P.  Part  of  its  circular  V.T.  will  be  the  arc 
fg.     Any  plane  whose  V.T.  is  a  tangent  to  this  arc  and  whose  H.T.  passes  through 


TANGENT   PLANES   COMMON   TO   THREE   GIVEN   SPHERES    97 

^  is  a  plane  at  45°  to  the  V.P.  because  it  is  tangent  to  a  right  cone  perpendicular 
to  the  V.P. 

We  now  have  two  cones  enveloping  the  same  sphere,  and  a  plane  containing 
both  apexes,  and  tangent  to  the  sphere,  will  contain  a  generator  of  each  cone. 
Such  a  plane  may  now  be  represented  by  drawing  its  H.T.  through  H,  tangent  to 
the  circular  H.T.  of  the  60°  cone  and  its  \'.T.  through  V,  tangent  to  the  circular 
V.T.  of  the  45°  cone.  These  traces,  being  traces  of  the  same  plane,  meet  in  the 
XY  at  R. 


EXERCISE   XXXVIII 

Find,  by  the  method  employing  enveloping  or  tangent  conea/to  a  sphere,  the  traces  of 
the  following  oblique  plane:  / 

{a)  At  70°  to  the  H.P.  and  at  55°  to  the  \'.P. 
{b)  At  35°  to  the  H.P.  and  at  75°  to  the  V.P. 
{c)  At  65°  to  the  H.P.  and  at  50°  to  the  V.P. 


TANGENT  PLANES  COMMON  TO  THREE  GIVfeN  SPHERES 

Section  27.  In  Fig.  97  is  illustrated  the  method  of  finding  tangent  planes 
common  to  three  unequal  spheres,  and  the  projections  of  the  tangent  points  on 
their  surfaces. 

The  three  spheres  .4,  B  and  C,  are  resting  on  the  H.P.,  so  that  enveloping 
cones  of  any  two  of  them  have  their  apexes  in  the  H.P.  The  enveloping  cone 
of  .1  and  B  has  its  apex  at  Hi,  and  one  enveloping  .1  and  C  has  its  apex  at  H-z- 
The  line  through  Hi  and  II2  is  therefore  the  H.T.  of  a  plane  tangent  to  the  three 
spheres,  and  by  passing  a  vertical  plane,  perpendicular  to  this  tangent  plane, 
through  the  centre  of  one  of  the  spheres,  say,  C,  and  rabatting  as  in  Fig.  95.  the 
tangent  point  cc'  is  obtained.  The  V.T  .of  the  tangent  plane  will  be  perpen- 
dicular to  the  elevation  of  the  radius  which  ends  in  c' . 

Just  as  the  radius  to  point  C  is  perpendicular  to  the  tangent  plane  found, 
so,  in  like  manner,  the  radius  lines  to  the  tangent  points  on  the  surfaces  of  A  and 
B  will  be  perpendicular  to  the  plane,  therefore  plans  of  these  radius  lines  made 
perpendicular  to  the  H.T.,  and  elevations  of  them  made  perpendicular  to  the  V.T. 
may  be  drawn.  These  must  be  limited  at  aa  and  bb'  by  generator  lines  on  the 
surfaces  of  the  cones,  one  from  H2  through  cc'  to  limit  the  radius  line  from  the 
centre  of  sphere  A,  and  another  from  this  point  aa'  found  on  sphere  A,  drawn  to 
III,  to  limit  the  radius  of  the  sphere  B  at  bb'.  The  elevations  of  the  points  may  be 
obtained  by  carrying  perpendiculars  across  A'l'  from  the  plans,  instead  of  using 
elevations  of  generator  lines.  Both  ways  are  shown  in  the  figure.  One  checks 
the  accuracv  of  the  other. 


98 


DESCRIPTIVE    GEOMETRY 


EXERCISE   XXXIX 

1.  Find  two  tangent  planes  to  three  spheres,  the  planes  not  to  pass  between  any  of  the 
spheres,  and  mark  the  projections  of  the  tangent  points  on  their  surfaces.  The  spheres  have 
their  centres  all  at  the  same  level,  and  these  centres  are  the  corners  of  an  equilateral  triangle 
2"  side,  with  no  side  parallel  to  the  V.P.     Diameters  of  the  spheres  if",  i\"  and  f "  respectively 

2.  Three  spheres  rest  on  the  H.P.  and  touch  each  other.  Their  centres  are  at  difTerent 
distances  from  the  V.P.  Their  diameters  are  if",  |"  and  f"  respectively.  Find  the  inclined 
plane  tangent  to  all  three  spheres  and  mark  the  projections  of  the  tangent  points. 


Fig.  97. 


When  the  centres  of  three  unequal  spheres  are  at  unequal  distances  from 
the  planes  of  projection  and  the  spheres  are  not  all  resting  on  the  H.P.,  then  the 
line  joining  the  apex  points  of  the  enveloping  cones  will  be  oblique,  and  the  problem 
becomes  more  involved.  If,  however,  the  line  joining  the  apex  points  be  considered 
in  relation  to  one  sphere  only,  then  it  becomes  a  matter  of  finding  the  planes  con- 
taining this  line  and  tangent  to  the  one  sphere.  It  will  be  seen  that  the  tangent 
planes  found  will  be  tangent  to  the  other  spheres,  the  apex  of  whose  common 
enveloping  cone  is  a  point  in  the  line.  The  tangent  points,  also,  having  been 
obtained  on  the  one  sphere,  the  tangent  points  on  the  others  may  be  found  as  in 
the  case  considered  in  Fig.  97. 

Let  ah,  a'h'  in  Fig.  98  be  an  inclined  line,  such  as  that  referred  to  above,  and 
cc'  the  centre  of  a  sphere  whose  projections  are  given.     Let  it  be  required  to  find 


TANGENT   PLANES  COMMON   TO    THREE   OIVEN   SPHERES 


99 


the  tangent  planes  to  this  sphere,  which  contain  the  Hne  AB.  The  order  of 
procedure  might  be  as  follows: — (i)  Find  the  traces,  II  and  V,  of  the  given  line. 
These  will  be  in  the  traces  of  the  planes  required.  (2)  Find  a  plane,  RST,  passing 
through  the  centre  of  the  sphere  and  perpendicular  to  the  line  AB.     (3)  By  the 


Fig.  98. 


help  of  a  plane  LMX,  containing  the  line  -1^,  and  intersecting  plane  RST,  find 
the  point  of  intersection  which  AB  makes  with  RST  at  />/>'.  (4)  Rabatte  the  point 
P  about  5rto  p2,  and  also  rabatte  the  centre  C  to  C2,  and  draw  the  rabatted  great 
circle  of  the  sphere  made  by  the  intersection  of  the  sphere  by  plane  RST  passing 
through  its  centre.     (5)  From  p-^  draw  the  rabatted  tangent  lines  to  this  great 


100  DESCRIPTIVE   GEOMETRY 

circle  and  find  their  H.T.'s  at  Ih  and  Ho.  (6)  Join  Hi  and  H2  to  p  and  so  obtain 
the  plans  of  two  lines  which  are  tangent  to  the  sphere  and  have  a  common  point 
P  in  the  given  line  AB.  (7)  The  planes  required,  tangent  to  the  sphere,  will  each 
contain  one  of  these  tangent  lines  and  the  given  Hne,  therefore  draw  the  traces 
of  the  required  planes  by  making  their  H.T's  at  HiH  and  H2H  respectively.  Their 
V.T.'s  will  pass  through  the  V.T.  of  the  given  line  AB,  viz.,  V.  (8)  The  projec- 
tions of  the  tangent  points  may  be  obtained  by  passing  them  over  from  their 
rabattements  at  J2  and  co  to  their  plans  at  d  and  e  respectively,  and  then  obtain- 
ino;  their  elevations  in  the  usual  way. 


EXERCISE   XL 

1.  Find  the  traces  of  the  tangent  planes  to  the  sphere,  of  2"  diameter,  which  has  its  centre 
i\"  from  both  planes  of  projection,  and  contains  a  line  whose  H.T.  is  2"  from  XY  and  2"  from 
the  plan  of  the  centre  of  the  sphere,  while  its  V.T.  is  3"  above  XY  and  if"  from  the  elevation 
of  the  centre  of  the  sphere.  Both  these  given  traces  are  to  the  right  of  the  centre  of  the  sphere. 
Also,  mark  the  projections  of  the  tangent  points  on  the  surface  of  the  sphere. 

2.  Arrange  three  spheres  of  difTerent  sizes  and  not  far  away  from  each  other,  placing  them 
at  different  levels.  Find  two  planes  tangent  to  all  the  spheres  and  mark  the  tangent  points  on 
the  spheres. 


CHAPTER  XIV 

SIMPLE  CASES  OF  IXTERPEXETRATIOXS  OF  SOLIDS,  AND  THE  DEVELOPMENTS 

OF  SURFACES 

Section   28.  The   matter   of   finding    intelligently    the    intersection    of    the 
surface  or  surfaces  of  one  solid  with  the  surface  or  surfaces  of  another,  depends 


IlG.  yy. 


chiefly  upon  recognizing  the  use  of  sectional  planes  or  of  surfaces  which  will  con- 
tain elements  or  generators  of  both  of  the  given  surfaces,  meeting  each  other  in 
points  common  to  both  given  surfaces.  The  solids  should  be  so  arranged  as  to 
provide  for  the  use  of  planes  which  will  readily  give  plans  and  elevations  of  the 
generators  or  elements,  and  their  intersection  points  common  to  both  surfaces. 
In  Fig.  99  is  shown  a  square  prism  penetrating  a  cylinder.     In  making  the 

101 


102 


DESCRIPTIVE   GEOMETRY 


projections  of  the  solids,  or,  rather,  such  parts  of  them  as  may  be  necessary, 
start  by  making  the  square  which  will  serve  as  the  rabattement  of  the  end  of 
the  square  prism. 

Inspection  of  the  figure  will  disclose  the  following  facts:— (i),  that  a  vertical 
section  plane  containing  the  axis  of  the  cylinder  contains  two  vertical  generators 
of  the  cylinder,  and  may  also  contain  the  edges  A  and  B  of  the  prism  which  has 
been  placed  parallel  to  the  V.P.;  (2)  that  the  elevations  of  the  points  of  intersec- 
tion of  these  two  edges  with  the  generators  of  the  surface  of  the  cylinder  are  a'a' 
and  b'b'  respectively;  (3)  that  similarly  c'c'  are  the  elevations  of  the  intersections 
of  the  edge  C  with  the  surface  of  the  cylinder;  (4)  that  for  other  points,  common 
to  both  surfaces,  straight  lines,  of  any  convenient  number,  and  placed  conveniently 


(i.e.,  not  necessarily  at  equal  distances  from  each  other)  to  run  parallel  to  the  V.P., 
are  drawn  on  the  surfaces  of  the  prism  at  i,  2,  3,  4,  first  in  plan  and  then  in  eleva- 
tion, by  taking  advantage  of  the  rabatted  square  end;  (5)  that  vertical  generator 
lines  on  the  surface  of  the  cylinder,  and  in  the  same  vertical  section  planes  as  the 
lines  on  the  prism,  pass  through  the  points  where  these  lines  i,  2,  3  and  4  cut  the 
curved  surface  of  the  cylinder.  These  are  then  projected,  resulting  in  the  eleva- 
tions of  points,  common  to  both  surfaces,  and  shown  at  d',  e',  f,  g' ,  etc.;  (6)  that 
these  elevation  points  must  be  joined  by  freehand  curved  lines  to  indicate  the 
elevation  of  the  common  section  Hues. 

In  Fig.  100  is  shown  what  is  called  the  development  of  the  surfaces  of  the  prism 
of  Fig.  99,  in  order  to  show  what  the  section  lines  on  those  surfaces  appear  like. 
The  long  lines  marked  ACBDA  are  set  up  at  distances  from  each  other  equal 
to  the  width  of  the  faces  of  the  prism.     Then  the  straight  lines  which  were  drawn 


SIMPLE   CASES    OF    IXTERPEXETRATIOXS   OF    SOLIDS       103 

upon  those  faces  in  Fig.  99  are  drawn  at  their  proper  distances  from  each  other, 
taken  from  the  rabatted  end  and  numbered.  The  distances  of  the  points  of  inter- 
section on  all  these  lines  and  upon  the  edges  ACBD  are  now  taken  from  the  eleva- 
tion in  the  figure,  and  joined  by  freehand  curves,  thus  marking  the  exact  way  in 
which  each  face  has  been  penetrated  by  the  surface  of  the  cylinder. 

In  a  similar  way  the  curved  surface  of  the  cylinder  may  be  "  develoi)cd  " 
and  the  intersection  lines  upon  its  surface  shown.     The  circumference  of  the  end 


Fig.  ioi. 


is  laid  out  as  a  straight  line,  and  pcri)cndiculars  are  made  from  it  to  represent 
generators.  These  are  cut  at  heights  taken  from  the  elevation  in  the  figure  and 
then  the  points  so  found  are  joined  by  curved  lines,  showing  the  e.xact  way  in 
which  the  surface  has  been  cut. 

A  further  illustration  of  development  is  given  in  Fig.  loi,  where  it  will  be 
seen  that  the  circular  base  of  the  solid  is  laid  out  as  well  as  the  cun'ed  surface, 
and  the  intersection  lines  made  on  them  are  displayed.  For  convenience  the 
curved  surface  is  divided  into  twelve  equal  parts.  One-twelfth  of  the  circum- 
ference of  the  base,  that  between  7  and  S  in  the  plan.  is.  by  an  approximate 


104  DESCRIPTIVE   GEOMETRY 

method,  found  to  be  equal  to  the  tangent  marked  76.  The  method  is,  to  make 
the  hne  from  point  7  through  the  centre  to  a,  equal  to  three  times  the  radius  of  the 
circle,  and  from  a  to  draw  ab  through  the  other  end  of  the  30°  arc  to  the  tangent 
line  from  7.  Since  all  the  division  lines  of  the  curved  surface  are  equal,  from  c  in 
the  elevation,  as  centre,  draw  the  arc  i,  2,  3,  etc.,  to  i,  marking  on  it  the  divisions, 
equal  to  the  line  jb  in  the  plan  drawing.  The  intersections  of  the  lines  on  the 
surface  of  the  cone  are  now  carried  across,  by  parallels  to  XY,  to  the  line  ci, 
and  from  that  line  carried  by  arcs  with  c  as  centre  to  their  development.  Between 
the  points  2  and  3  the  circumference  of  the  base  is  cut  by  the  plane  RST,  and  at 
this  point,  therefore,  in  the  development,  draw  a  tangent  circle,  so  that  the 
intersection  of  the  base  from  this  point  may  be  marked  on  the  circle  representing 
it  in  the  development. 


EXERCISE  XLI 

1.  Find  the  elevation  of  the  intersection  lines,  made  by  the  surfaces  meeting  each  other 
of  a  square  prism,  2"  short  edge,  standing  vertically  with  a  vertical  face  at  30°  to  the  V.P.,  and 
another  square  prism,  i|"  short  edge,  with  long  edges  horizontal  and  parallel  to  the  V.P.  A 
face  of  this  prism  is  at  20°  to  the  H.P.,  so  that  both  of  the  diagonals  of  the  end  of  it  are  inclined, 
the  one  at  65°  and  the  other  at  25°,  to  the  H.P.  Arrange  it  so  that  three  of  the  vertical  edges  of 
the  first  prism  pass  -through  the  second  prism,  and  one  of  the  vertical  edges  of  the  first  passes 
through  the  axis  of  the  second.  (N.B. — In  working  out  the  result,  it  must  be  seen  that  the  vertical 
edges  of  the  first  prism  penetrate  faces  of  the  second  prism  in  points  through  which  should  be 
drawn,  on  the  faces  of  the  second  prism,  hnes  parallel  to  the  long  edges.  The  elevations  of 
these  should  then  be  obtained.) 

2.  A  right  cylinder,  2"  diameter,  has  its  axis  vertical,  and  a  square  prism,  with  one  diagonal 
of  end  horizontal  and  at  70°  to  the  V.P.,  has  its  axis  inclined  at  20°  to  the  H.P.  The  axis  of 
the  prism  passes  through  the  axis  of  the  cylinder.  The  edge  of  end  of  the  prism  is  ij"  long. 
(N.B. — It  is  suggested  that  in  setting  up  these  solids  the  prism  should  be  arranged  correctly 
first,  in  relation  to  the  planes  of  projection,  and  the  cylinder  placed  in  position  after  this  has 
been  done.) 

3.  Two  right  circular  cylinders  interpenetrate  each  other.  Let  one  be  placed  with  its  axis 
vertical,  its  diameter  being  2",  and  let  the  other,  with  diameter  if",  have  its  axis  inclined  at 
20°  to  the  H.P.  and  parallel  to  the  V.P.,  is"  farther  away  from  the  V.P.  than  the  axis  of  the 
other  cylinder.  Show  the  elevation  of  the  intersection  curve  made  by  their  surfaces.  Make 
also  a  development  of  the  curved  surface  of  the  2"  cylinder  to  show  the  intersection  line  upon 
it. 


MORE  DIFFICULT  CASES  OF  INTERPENETRATION  OF  SOLIDS,  AND  THE 
PROJECTION  OF  THE  INTERSECTIONS  OF  THEIR  SURFACES 

Section  29.  When  the  intersecting  surfaces  are  those  of  the  cone  and  the 
sphere,  then  all  that  is  necessary,  in  order  to  obtain  the  projection  of  points 
common  to  both  surfaces,  is  to  make  use  of  planes  which  give  circular  sections  of 
the  cone,  and  pass  through  the  sphere.     If  the  circular  sections  of  the  sphere, 


MORE  DIFFICULT  CAbES  OF  INTEKPEXETRATIOX  OF  SOLIDS    lu5 

by  these  planes,  intersect  the  circular  sections  of  the  cone  by  the  same  planes, 
then  the  points  of  intersection  are  points  on  both  surfaces. 

Another  method  of  determining  the  intersection  of  the  surfaces  of  these  soHds. 
is  by  the  use  of  intersecting  spherical  surfaces  instead  of  intersecting  planes,  and 
depends  on  the  following  facts: — 

(i)  That  a  sphere  whose  centre  is  in  the  axis  of  a  right  circular  cone  will 
intersect  the  surface  of  that  cone,  if  it  intersects  at  all.  in  a  circle  perpendicular 
to  the  axis  of  the  cone; 


Fig.  io: 


Fig.  103. 


(2)  That  if  a  sphere  intersects  another  sphere  the  intersection  of  their  sur- 
faces is  the  circumference  of  a  circle,  the  plane  of  which  is  perpendicular  to  the  line 
joining  their  centres;  and 

(3)  That  if  a  sphere,  with  its  centre  in  the  axis  of  a  right  circular  cone  inter- 
penetrates at  the  same  time  another  sphere,  then  the  intersection  of  the  circular 
sections  it  makes  with  both  solids  will  be  points  common  to  the  surfaces  of  both 
solids. 

Illustrations  of  the  use  of  both  methods  referred  to  are  given  in  Figs.  102 
and  103.     In  Fig.  102,  a  right  circular  cone  with  its  axis  vertical,  and  a  sphere 


106 


DESCRIPTIVE   GEOMETRY 


whose  centre  is  cc',  are  arranged  so  that  their  surfaces  intersect.  The  centre 
of  the  sphere  is  not  necessarily  at  the  same  distance  from  the  V.P.  as  the  axis  of 
the  cone.  By  the  use  of  horizontal  section  planes  a  number  of  points  will  be 
found  in  plan,  where  the  circular  sections  of  the  solids  intersect.  Their  elevations 
are  readily  discovered,  and  by  joining  the  points  by  freehand  curves,  the  projec- 
tions of  the  section  line  or  lines,  are  obtained. 

In  Fig.  103,  concentric  sectional  spheres,  having  their  common  centre  at  the 
apex  of  the  cone,  are  represented  a.t  A,  B,  C,  etc.     Since  the  centre  of  the  given 


Fig.  104. 


sphere  is  at  the  same  distance  from  the  V.P.  as  the  axis  of  the  given  cone,  the 
sections  the  given  cone  and  sphere  make  with  the  concentric  spheres  will  appear  in 
elevation  as  straight  lines,  intersecting  each  other  in  the  elevations  of  points  com- 
mon to  their  surfaces.  These  points  may  readily  be  found  in  plan  in  the  plans 
of  the  circular  sections  of  the  cone.  For  example,  the  cone  intersects  the  sphere 
A  in  the  circle  whose  elevation  is  a'a',  and  the  given  sphere  intersects  the  sphere 
A  in  the  circle  whose  elevation  is  a' 2a' 2.  Where  these  cross  each  other  is  the 
point  a'z,  which  is  the  elevation  for  two  points  a^  and  as  in  plan. 


MORE  DIFFICULT  CASES  OF  INTERPEXETRATIOX  OF  SOLIDS     107 

In  Fig.  104  is  shown  how  best  to  deal  with  the  problem  of  linding  the  plan 
and  elevation  of  the  intersection  of  the  surface  of  a  cone  with  the  surface  of  a 
cylinder,  when  the  two  solids  interpenetrate. 

The  c>linder  is  placed  with  its  a.xis  parallel  to  the  V.P.,  and  a  secondary  ele- 
vation plane  is  represented,  perpendicular  to  the  ordinary  V.P.,  with  its  XY 
marked  X2Y2  in  the  figure.  Upon  this  secondary  vertical  plane'  is  drawn  an 
end  view  of  the  cylinder,  and  another  view  of  the  cone. 

Inclined  planes  perpendicular  to  this  secondary  vertical  plane,  and  con- 
taining generators  of  the  cylinder  and  of  the  cone,  are  represented  with  their  H.T.'s 
at  A,  B,  C,  etc.,  and  their  V.T.'s,  passing  through  the  secondary  elevation  of  the 
apex  of  the  cone  at  c".  One  generator  of  the  cone  and  two  generators  of  the 
cylinder  are  contained  in  the  plane  A .  The  intersections  made  by  the  generators 
result  in  two  points  whose  elevations  are  a  and  a'2.  Plane  B  contains  two  genera- 
tors of  each  solid  and  consequently  four  points  common  to  the  two  surfaces  are 
found;  and  so  on.  After  sufTicient  points  are  found  the  freehand  curves  may  be 
drawn  through  them.  In  the  figure,  the  elevation  has  been  completed,  in  order 
to  emphasize  the  fact  that  it  is  better  to  work  it  out  first.  The  plan  may  now 
jjc  obtained,  either  by  projecting  from  the  elevations  of  the  points  on  to  the  plans 
of  the  generators  of  the  cone,  or  by  making  plans  of  the  generators  of  the  cylinder 
to  cross  the  plans  of  the  generators  of  the  cone  in  the  points  required.  This  latter 
method  will  be  necessary  in  the  case  of  the  plans  to  be  marked  a  and  a-z. 


EXERCISt:   XLII 

1.  Find  plan  and  elevation  of  the  intersection  of  the  surfaces  of  a  right  circular  cone  and  a 
sphere,  by  using  the  method  involving  spherical  section  surfaces.  The  cone  has  a  base  2 J" 
diameter,  in  the  H.P.  and  a  height  2^".  The  sphere,  diameter  2",  rests  on  the  H.P.,  and  has 
its  centre  in  the  curved  surface  of  the  cone. 

2.  Find,  by  the  method  of  horizontal  section  planes,  the  plant  and  elevation  of  the  inter- 
section line,  when  a  cone  with  25"  circle  for  its  H.T.,  and  with  an  axis  2§"  long  from  the  centre 
of  its  H.T.  to  its  ape.x,  inclined  at  70°  to  the  H.P.,  interpenetrates  with  a  sphere,  2"  diameter, 
resting  on  the  H.P.  and  having  its  centre  in  the  shortest  line  from  the  apex  to  the  H.P..  on  the 
surface  of  the  cone. 

3.  A  right  circular  cylinder.  2"  diameter,  lies  with  one  of  its  generators  in  the  H.P.,  parallel 
to  the  \'.P.  A  cone,  whose  H.T.  is  a  circle  3"  diameter,  has  the  centre  of  this  circle  in  the 
plan  of  the  axis  of  the  cylinder.  The  axis  of  the  cone  is  4"  long,  and  has  a  plan  J"  long  at  45° 
to  XY.     Find  the  plan  and  elevation  of  the  intersection  of  the  surfaces  of  the  two  solids. 


To  obtain  the  projections  of  the  surface  intersections  when  inclined  cones  and 
cylinders,  not  necessarily  right  circular  ones,  interpenetrate,  it  is  necessary  to  make 
use  of  section  planes  which  will  have  in  them  straight  generator  lines  on  both  sur- 
faces. Hence,  when  the  surfaces  of  cones  are  cut  by  those  of  other  cones,  or  of  cylin- 
ders, the  apex  points  of  the  cones  must  be  in  the  section  planes  made  use  of;  and  in 


108 


DESCRIPTIVE   GEOMETRY 


the  case  of  cylinders,  their  axes  must  be  parallel  to  the  section  planes.  Thus,  in  Fig. 
105,  where  two  cylinders  are  represented  whose  given  H.T.'s  are  circles  and  whose 
axes  are  directed  as  indicated  by  the  plans  and  elevations  of  generators  of  their 


Fig.  105. 


surfaces,  it  is  necessary  to  take  any  generator  of  one  cylinder,  or  a  parallel  to  it 
as  ab,  a'b' ,  and  from  some  point  in  that  line  to  draw  another  line,  say  ac,  a'c', 
parallel  to  the  generators  of  the  second  cylinder.  The  H.T.'s  of  these  two  lines 
are  h  and  c  respectively,  and  the  plane  of  the  two  lines  has  the  line  HH  for  its  H.T. 
All  the  generators  of  both  surfaces  are  parallel  to  this  plane,  and  hence  any  plane 


MORE  DIFFICULT  CASES  OF  IXTERPENETRATIOX  OF  SOLIDS     109 

made  parallel  to  this  one,  and  having  its  H.T.  crossing  the  circular  H.T.'s  of  the 
given  cyhnders  will  therefore  contain  generators  of  the  surfaces  of  both  cylinders. 
The  points  in  which  these  generators  cut  each  other  are  points  in  the  intersection 
line  of  the  surfaces,  and  the  plan  and  elevation  of  it  can  in  this  way  be  found,  as 
shown.  It  will  be  found  an  advantage  to  arrange  tangent  planes  to  commence 
with.  These  in  Fig.  105,  are  marked  i  and  2  respectively,  and  it  will  be  there  seen 
that  a  strip  on  each  of  the  cylinders  is  not  pierced  by  the  surface  of  the  other,  and 
the  curves  will  be  turned  back  from  these  strips  accordingly. 


Fio.  106. 


In  Fig.  106  two  cones  are  represented.  A  hne  through  the  ape.\  points 
will  be  contained  in  the  planes  which  have  generators  of  both  surfaces.  Hence 
the  V.T.  and  the  H.T.  of  this  line  must  be  found,  as  at  V  and  Z7,  and  the  traces  of 
planes  passed  through  them.  Those  planes,  such  as  VRII,  which  contain  genera- 
tors of  both  cones,  will  give  points  common  to  both  surfaces.  A  number  of  such 
points  must  be  found  in  order  to  obtain  the  projections  of  the  cur\-e  or  curves. 

In  Fig.  107,  a  cylinder  and  a  cone  are  represented,  and  here  it  will  be  seen 
that  since  the  apex  of  the  cone  must  be  in  all  the  section  planes  that  are  of  sersice 
in  finding  the  intersection,  and  these  section  planes  must  contain  generators  of 


no 


DESCRIPTIVE   GEOMETRY 


the  cylinder  also,  therefore  they  must  contain  a  line,  through  the  apex  point, 
parallel  to  the  generators  of  the  cylinder.  Having  drawn  this  line  and  obtained 
its  traces,  as  at  V  and  H,  it  is  only  necessary  then  to  draw  traces  of  section  planes 
through  V  and  H  which  will  cross  the  H.T.  of  the  cylinder  and  the  V.T.  of  the 


Fig.  107. 


cone,  as  the  plane  VRH  does,  and  obtain  plans  and  elevations  of  generators  of 

both  surfaces,  as  they  cross  each  other,  thus  finding  projections  of  section  points. 

Other  simple  cases  of  interpenetration,  not  requiring  for  their  solution  any 

new  principle,  or  serious  change  of  method,  will  doubtless  suggest  themselves. 

EXERCISE   XLIII 


Work  to  completion,  in  plan  and  elevation,  the  cases  of  interpenetration  shown  in  Figs.  106 
and  107. 


SURFACES   OF   REVOLUTION,   AND   THE  SCREW   THREAD     U; 


SURFACES  OK  RE\OLLTIOX,  AND  THE  SCREW  THREAD 

Sectiox  30.  Thus  far  we  have  dealt  with  three  different  surfaces  of  revolu- 
tion, the  cylinder,  the  cone  and  the  sphere.  The  first  is  formed  by  revolving  a 
straight  line  round  an  axis  of  rotation  while  maintaining  a  parallel  attitude  in 
relation  to  it.  The  second  is  formed  by  revolving  a  straight  line  round  an  axis 
of  rotation,  the  revolving  line  having  one  point  of  it  in  the  axis  of  rotation.  The 
third  is  formed  by  revolving  a  semicircle,  making  use  of  its  diameter  as  the  axis 
of  rotation.  Others  will  now  be  considered,  namely,  the  hyperboloid  and  the 
helicoid. 


The  hyperboloid  of  revolution  is  obtained  as  the  result  of  a  straight  line  revolv- 
ing about  an  axis  of  rotation,  but  not  in  the  same  plane  with  it.  Each  point  in 
the  revolving  line  travels  in  the  circumference  of  a  circle  for  its  locus,  the  plane 
of  which  is  at  right  angles  to  the  axis.  This  will  be  noted  in  Fig.  io8,  where  c, 
c'c'  is  the  axis  of  rotation,  and  ah.  ah'  is  the  line  revolving.  .\ny  points  taken 
in  this  line  revolve  in  paths  which  are  the  circumferences  of  circles  whose  planes 
are  horizontal,  the  axis  being  vertical,  and  on  joining  the  elevations  of  the  extreme 
limits  to  which  these  points  move  from  one  side  to  the  other,  the  freehand  cur\-e 
gives  the  projection  of  the  contour  of  the  surface  of  revolution  generated,  namely, 
the  hyperboloid  of  revolution. 

The  helicoid.  or  surface  of  revolution  generated  by  a  straight  line  moving 
so  that  any  point  in  it  travels  in  a  helicoidal  line  or  helLx,  as  a  spiral,  while  the 


112 


DESCRIPTIVE   GEOMETRY 


line  is  not  in  the  same  plane  with  the  axis,  but  is  constantly  maintaining  the  same 
attitude  in  relation  to  it,  is  used  in  the  making  of  screw-threads  on  bolts,  etc. 


Fig.  iio. 


In  projecting  screw-threads,  the  curved  edge  lines  of  the  threads  are  the 
paths  of  points  in  the  generating  lines  of  hehcoids,  and  may  be  thought  of  as  lines 


SURFACES   OF   REVOLUTION,  AND   THE   SCREW   THREAD     113 

drawn  on  the  surfaces  of  cylinders  or  drums.  The  development  of  such  a  cylin- 
drical surface  of  a  bolt,  showing  the  thread  line,  would  give  that  line  as  a  straight 
one,  as  ac  in  the  Fig.  109.  The  Hne  ab  is  the  length  of  the  circumference  of  the 
circular  section,  and  the  distance  be  represents  the  pitch  or  distance  travelled  in 
the  direction  of  the  axis  during  one  revolution,  ac  is  the  straight  line  represent- 
ing the  helicoidal  line  developed  as  from  the  surface  of  a  cylinder.  These  curved 
helicoidal  lines  may  be  projected  on  to  the  V.P.,  and  in  this  way  a  screw-thread  is 
represented. 

As  in  illustration,  in  Fig.  no,  let  ABB2  be  the  triangular  section,  made 
by  a  plane  which  has  in  it  the  axis  of  the  bolt,  of  a  screw-thread  to  be  pro- 
jected, when  the  large  circle  in  the  plan,  represents  the  section  of  the  bolt.  From 
B  to  B2  is  the  pitch  of  the  thread.  Two  helicoidal  lines  must  be  found,  one,  on 
the  outside  cylindrical  surface,  and  passing  through  the  point  A,  and  the  other 
on  an  inner  cylinder  or  drum  upon  which  is  situated  the  point  B.  Notice  that 
the  pitch  is  the  same  for  both  lines. 

For  convenience,  30°  divisions  of  the  circular  plan  are  made,  therefore  twelve 
in  all,  and  for  each  30°  division  over  which  the  line  moves  it  rises  ,Vt  of  the  pitch. 
For  a  right-hand  screw  the  thread  runs  upwards  in  the  direction  of  the  arrow- 
headed  line. 

For  a  thread  of  square  section  there  will  be  four  helicoidal  lines  to  draw.  In 
such  a  case  if  the  pitch  be  f ",  the  edge  of  the  square  section  will  be  |". 


EXERCISE   XLI\' 

1.  Find  the  elevation  of  a  screw-thread  on  a  vertical  bolt  2^"  diameter,  when  the  section 
of  the  thread  by  a  plane  including  the  a.xis,  is  an  equilateral  triangle.  The  thread  to  be  single, 
and  pitch  |". 

2.  Find  the  projection  of  a  bolt  screw-thread  with  square  section.  Diameter  of  bolt  3", 
pitch  f . 

3.  Find  the  surfaces  of  revolution  generated  by  the  edges  of  a  cube  made  to  rotate  on  one 
of  its  solid  diagonals,  i.e.,  on  a  diagonal  passing  through  its  centre.     Edge  2". 


CHAPTER  XV 

RADIAL  PROJECTION.     PERSPECTIVE  PROJECTION 

Section  31.  Things  are  made  visible  to  us  by  means  of  light,  and  in  order 
to  obtain,  in  the  eye,  a  picture  of  anything,  rays  of  light,  which  travel  in  straight 
lines,  must  proceed  from  the  thing  looked  at,  to  the  eye.  The  retina,  at  the  back 
of  the  eye,  is  the  picture  surface,  a  curved  one,  receiving  the  projection  or  view 
of  the  object.  This  projection  or  view  is  obtained  by  a  process  called  Radial 
Projection.  It  is  the  same  as  that  by  which  a  plane  surface,  the  plate  or  flat 
film,  in  a  camera,  receives  a  picture. 

In  Orthographical  Projection,  by  which  we  have  obtained  plans  and  eleva- 
tions of  things,  the  projectors  are  at  right  angles  to  the  projection  plane.  In 
Radial  Projection  the  projectors  are  inchned  to  the  projection  plane.  Radial 
Projection  is  used  in  different  ways  for  the  projection  of  maps,  etc.,  but  is  most 
commonly  used,  by  Architects  and  others,  in  what  is  called  Perspective  Projection. 

In  Perspective  Projection  a  single  projection  plane  is  used,  and  is  situated 
between  the  object  and  the  eye.  It  can  be  seen  through,  by  the  spectator,  as 
he  looks  at  the  object.  The  view  or  projection  of  the  object  is  caught  by  this 
projection  plane,  or  Picture  Plane  as  it  is  called,  which  is,  in  attitude,  at  right 
angles  to  the  direction  of  vision. 

Since,  in  a  natural  way,  and  ordinarily,  a  person's  sight  is  directed,  as  he  stands, 
in  a  horizontal  direction,  the  Picture  Plane,  in  Perspective  Projection,  is  arranged 
as  a  vertical  plane,  with  a  point,  perpendicularly  opposite  the  eye,  marked  upon 
it,  as  the  Centre  of  Vision. 

The  eye-level  exactly  agrees  with  what  one  sees  in  the  distance  as  the  hori- 
zon, and  is  represented  on  the  Picture  Plane  by  a  horizontal  Hne,  in  which  the 
Centre  of  Vision  point  is  situated.  This  line,  representing  the  eye-level  and  at 
the  same  time  the  horizon,  is  called  the  Horizon  Line,  and  parallel  to  it,  on 
the  Picture  Plane,  another  horizontal  Hne  is  drawn,  at  a  distance  below  the 
Horizon  Line  equal  to  the  distance  of  the  spectator's  eye  from  the  ground  and  is 
consequently  called  the  Ground  Line. 

The  distance  from  the  Horizon  Line  to  the  Ground  Line  is  decided  upon 
for  each  case,  and  should  be  such  as  will  best  serve  the  purpose  m  viewing  the 
object  to  be  represented.  Drawings,  of  course,  are  necessarily  made  strictly  to 
some  chosen  scale — |"  to  the  foot,  ^"  to  the  foot,  half  size,  etc.,  as  may  be  suitable. 

The  distance  between  the  object  and  the  spectator  is  variable,  and  may  be 
any  distance  chosen.     The  appearance  of  the  object  depends  very  much  upon  the 

114 


RADIAL   PROJECTION.     PERSPECTIVE   PROJECTION         115 

distance  chosen.  This  distance  being  settled,  it  is  then  necessary  to  fix  the  posi- 
tion, between  the  object  and  the  spectator,  of  the  Picture  Plane,  and  upon  this 
will  depend  the  size  of  the  drawing  or  projection  of  the  object. 

Let  the  student  imagine  the  vertical  transparent  Picture  Plane,  between 
his  eye  and  the  object,  and  he  will  realize  that,  as  the  plane  is  placed  nearer  to  the 
object  the  view  of  it  which  he  could  trace  on  the  plane  will  be  larger,  and  as  he 
brings  the  Picture  Plane  towards  the  eye,  the  view  of  the  object,  traced  on  the 
plane,  will  be  smaller. 


X     Plan  of  Picture  Piano 


Xz 


\<-V 


S*Plan  of  Spectator's  Eye 


Fig.  III. 


This  and  other  matters  will  be  better  understood  if  reference  is  made  to  Fig. 
Ill,  where  the  proper  arrangement  for  perspective  projection  is  shown  in  plan. 
It  will  be  seen  that  the  spectator  is  at  a  distance  OS  from  the  object,  and  that 
A'l',  representing  the  vertical  picture  plane,  at  right  angles  to  the  direction  of 
vision  rei)resentcd  by  the  direction  of  the  line  OS,  is  placed  not  far  away  from 
the  object  in  relation  to  the  distance  of  the  spectator  from  it,  and  consequently, 
the  size  of  the  picture  of  the  object  obtained  by  the  interception  of  the  rays,  from 
the  object  to  the  eye.  by  the  plane  at  .VI'.  will  be  fairly  large.     On  the  other 


116  DESCRIPTIVE   GEOMETRY 

hand,  if  the  picture  plane  be  placed  so  as  to  have  its  plan  at  X2Y2,  that  is,  rather 
close  to  the  eye  in  relation  to  the  distance  of  the  object  from  it,  then  the  picture 
obtained  by  the  interception  of  the  rays  from  the  object  as  they  proceed  to  the 
eye,  will  be  correspondingly  small.  The  point  C  is  the  plan  of  what  was  spoken 
of  as  the  Centre  of  Vision,  and  is  commonly  marked  CV. 

If  a  vertical  plane  be  imagined  as  containing  the  horizontal  line  of  sight 
or  direction  of  vision,  the  line  SC,  then  anything  to  the  left  of  this  vertical  plane, 
as,  for  instance,  the  corner  A  of  the  object,  is  said  to  be  to  the  left  of  the  spectator, 
and  anything  to  the  right  of  this  same  vertical  plane,  as  5,  is  said  to  be  to  the  right 
of  the  spectator.  The  whole  object  might  be  to  the  left,  or  to  the  right,  of  the 
spectator,  and  of  course  that  would  mean  that  the  plan  of  it  would  have  to  be 
placed  to  the  left,  or  right,  of  the  line  5C,  accordingly. 

In  order  to  make  a  perspective  projection  or  drawing,  the  essential  things  are: — 

(i)  The  placing  of  the  spectator  S,  and  the  Une  of  sight  SC. 

(2)  The  placing  of  the  object,  represented  by  plan,    at  its  proper  distance 

from  the  spectator,  measured  in  the  direction  of  the  line  of  sight,  and 
at  its  proper  distance  to  the  right  or  left  of  the  spectator. 

(3)  The  placing  of  the  picture  plane,  represented  by  XY,  and 

(4)  The  decision  as  to  the  height  of  the  eye  above  the  horizontal  plane  of 

the  ground,  on  which,  or  in  relation  to  which,  the  object  is  placed. 

To  represent  the  two  lines  Ground  Line  and  Horizon  Line,  two  lines,  so 
named,  are  drawn  at  their  proper  distances  apart,  across  the  paper  parallel  to  XY, 
and  at  any  convenient  place  between  the  point  5  and  the  line  XY,  as  in  Fig.  112. 

Before  proceeding  to  discuss  specific  cases  of  perspective  projection,  certain 
facts  which  have  a  bearing  on  the  intelligent  understanding  of  the  method  adopted, 
must  be  realized.     These  are  as  follows: — 

(i)  All    parallel    lines    receding    from    the    spectator    appear    to    converge, 
regardless  of  attitude  or  direction. 

(2)  All  horizontal  lines  regardless  of  level,  when  receding  from  the  spectator 

approach  the  horizon,  and  when  produced  to  infinity  lose  themselves 
on  the  horizon  in  points  commonly  spoken  of  as  vanishing  points,  and 

(3)  Any  number  of  parallel  horizontal  lines,  regardless  of  levels  or  direction, 

have  the  same  vanishing  point  in  the  horizon. 

Consequently,  therefore,  lines  on  the  ground  plane,  or  other  horizontal 
lines  below  the  level  of  the  eye,  if  receding  from  the  spectator,  will  be  represented 
by  lines  rising  in  the  projection  from  their  near  ends  towards  the  eye-level  line 
which  represents  the  horizon,  and,  similarly,  horizontal  receding  lines  above  the 
level  of  the  eye  will  be  represented  by  lines  drawn  downwards  from  their  near  ends 
towards  the  eye-level  line  where  their  vanishing  points  are. 

Consequently,  also,  horizontal  lines  that  are  perpendicular  to  the  projection 
plane,  that  is,  in  the  same  direction  as  the  line  of  sight,  wherever  they  are,  will 


RADIAL   PROJECTION.     PERSPECTIVE   PROJECTION        117 

have  their  vanishing  point  at  C,  or  CA' .,  in  the  horizon  Hne,  that  is,  at  the  Centre 
of  Vision. 

An  illustration  is  shown  in  Fig.  112,  where  a  vertical  shaft  standing  centrally 
on  a  square  block  is  represented.     The  plan  of  this  group  is  placed  a  little  to  the 


Fig.  112. 


left  of  the  spectator,  measured  to  the  left  of  the  line  of  sight,  and  is  at  a  distance, 
measured  along  the  line  of  sight,  from  the  spectator  whose  position  is  marked  5. 
The  picture  plane,  that  is,  the  projection  plane,  is  placed  vertically  at  AT,  which 
is  its  plan,  and  is  then  represented  again,  lower  down  on  the  paper,  with  .VI'  or 
Ground  Line  parallel  to  the  original  position  and  the  eye-level  line  or  Horizon 


118  DESCRIPTIVE   GEOMETRY 

Line  represented  at  the  height  decided  upon  for  the  eye.  In  the  figure  the 
elevation  of  the  group  is  shown,  exactly  opposite  the  plan  of  it,  on  this  plane,  and 
in  this  way  the  true  heights  on  the  picture  plane,  where  horizontal  lines  through 
the  points  of  the  group,  and  perpendicular  to  the  plane,  meet  it,  are  obtained. 

To  obtain  the  Perspective  Projection  it  is  now  only  necessary  to  represent 
all  the  horizontal  lines  that  are  really  perpendicular  to  the  Picture  Plane,  as  pro- 
ceeding to  the  Centre  of  Vision,  C.V.,  and  drop  perpendiculars  to  them  from 
the  points  in  the  plan  XY  of  the  picture  plane,  where  rays  from  the  points,  to  the 
eye,  are  intercepted  by  the  picture  plane.  In  the  figure  most  of  these  perpen- 
diculars are  only  started,  as  dotted  lines,  not  carried  all  the  way,  in  order  to  avoid 
confusion. 

Fig.  112  illustrates  what  is  often  spoken  of  as  Parallel  Perspective,  since  the 
lines  necessary  for  it  are  either  parallel  to  the  picture  plane  or  parallel  to  the  Line 
of  Sight. 

N  B. — Since  the  view  or  projection  is  not  influenced,  except  in  size,  by  the 
distance  of  the  picture  plane  from  the  spectator,  and  the  things  represented  are 
often  large  things  drawn  to  a  small  scale,  the  picture  plane  is  arranged  as  if  near 
the  object  in  order  to  get  a  reasonably  large  drawing  for  the  projection. 

EXERCISE  XLV 

1.  Find  the  perspective  projection  of  a  box  with  a  lid.  Its  measurements  are  2  ft.  X3  ft. 
and  it  is  i|  ft.  high.  Thicknesses  may  be  omitted.  Let  the  picture  plane  be  10  ft.  from  the 
spectator  and  the  box  2  ft.  beyond  the  plane,  or  2  ft.  "  into  the  picture  "  as  it  is  said,  and  the 
nearest  corner  4  ft.  to  the  left.  Let  one  end  be  parallel  to  the  picture  plane  and  the  front  of 
the  box  be  on  the  near  side.     Let  the  lid  be  opened  to  an  angle  of  45°. 

2.  Make  a  perspective  projection  of  a  double  cross,  that  is  having  four  arms  at  right  angles 
to  each  other,  with  a  pedestal  and  shaft  similar  to  those  in  Fig.  112.  Let  the  spectator's  eye 
be  5  ft.  above  the  ground  and  the  total  height  of  the  group  10  ft.,  the  pedestal  being  i^  ft.  thick 
and  4  ft.  square,  one  edge  of  the  square  being  parallel  to  the  picture  plane  and  3  ft.  into  the 
picture  (i.e.,  beyond  the  picture  plane).  Make  the  shaft  and  crossbars  i  ft.  square  in  section, 
each  arm  being  i|  ft.  long,  and  the  distance  from  pedestal  to  crossbars  5  ft.  Let  the  nearest 
corner  of  the  pedestal  be  3  ft.  to  the  right  of  the  spectator  and  the  distance  of  the  picture  plane 
from  the  spectator  12  ft. 

3.  Make  a  perspective  projection  of  any  familiar  object  such  as  a  book,  a  hut,  a  boat-house, 
a  table,  or  any  simple  thing  arranged  in  parallel  perspective. 


PERSPECTIVE  PROJECTION,  Continued 

Section  32.  We  have  seen  that  the  vanishing  point,  in  the  horizon,  for  hori- 
zontal lines  perpendicular  to  the  picture  plane,  is  the  same  as  that  for  the  line  of 
sight  or  line  from  the  eye  perpendicular  to  the  picture  plane,  and  marked  C.V.^ 
the  central  vanishing  point,  or  Centre  of  Vision. 

We  shall  now  consider  the  determination  of  vanishing  points  for  lines  that 
are  not  perpendicular  to  the  picture  plane.     For  a  horizontal  line,  the  vanish- 


PERSPECTIVE   PROJECTION,   CONTINUED 


119 


ing  point  must  be  in  the  horizon,  and  must,  for  our  purposes  of  projection,  be 
represented  by  a  point  on  the  horizon  line  drawn  on  the  picture  plane.  This  point, 
the  given  line's  vanishing  point,  will  be  the  same  as  for  a  line  proceeding  from 
the  eye  of  the  spectator  parallel  to  the  given  line.  Hence  the  method  shown  in 
Fig.  113,  namely,  from  S  draw  a  line  parallel  to  the  plan  of  the  given  horizontal 
line,  ab,  to  meet  XV,  the  projection  plane,  in  a  point  whose  plan  is  V,  and  whose 
place  in  the  Horizon  Line  on  the  picture  plane  is  V.P. 

Since  the  line  ab  and  the  parallel  to  it  SV  are  in  the  same  plane,  an  oblique 
plane,  it  will  be  realized  that  the  trace,  on  the  picture  plane,  of  the  line  ab,  viz., 


xi'      Plan  of  Pi  Mro  riano  V         Y 


Horizon  Line  /  tV 


I  ..  I 


X 


Fig.  ii.v 


the  point  whose  plan  is  v,  joined  to  the  trace  of  the  line  5F  at  the  point  whose 
plan  is  r,  will  give  the  intersection  of  the  oblique  plane  with  the  picture  plane. 
This  is  shown  as  the  line,  across  the  picture  plane,  marked  v'V ,  and  so  it  will 
be  realized  that  the  perspective  projection  of  the  line  ab,  a  line  in  the  oblique 
plane,  as  seen  from  the  point  5,  a  point  in  the  same  plane,  by  radial  projection  in 
that  oblique  plane,  must  be  a  foreshortened  line  lying  in  the  intersection  line 
v'V  in  the  projection  plane.  The  plans  of  the  rays  are  directed  from  a  and  b  to 
S  in  the  drawing,  and  are  seen  to  meet  the  plan  of  the  projection  plane,  AT.  at 
points  from  which  vertical  projectors  are  drawn  to  determine  the  perspective 
projection  ah' .     The  point  in  the  Horizon  Line  marked  \"  is  the  vanishing  point 


120 


DESCRIPTIVE   GEOMETRY 


for  the  line  AB  and  for  all  other  lines  parallel  to  it>-It  is  therefore  also  marked 
V.P.  A  second  hne,  horizontal,  and  immediately  above  the  hne  AB,  is  shown 
in  projection  at  a'2^'2-  Its  vertical  trace  in  the  picture  plane  is  y'2.  The  inter- 
section, of  the  plane  of  this  second  line  and  the  eye  point,  with  the  picture  plane, 
is  the  trace  line  v'2V',  and  since,  as  before,  the  rays  of  light  from  thi§  upper  line 


Fig.  114. 


travel  to  the  eye  in  the  obhque  plane  whose  intersection  with  the  picture  plane 
is  v'oV,  the  representation  of  the  line  in  perspective  is  inclined,  and  is  found 
to  be  a'2&'2. 

In  Fig.  114,  an  illustration  is  given  involving  the  use  of  vanishing  points  other 
than  the  central  vanishing  point.     These  are  marked  V.Pi  and  V.Po. 

It  will  be  noticed  that  as  there  are  no  lines  in  the  object  perpendicular  to  the 
Picture  Plane,  the  C.V.  does  not  come  into  use  as  a  vanishing  point  in  this  case. 


PERSPECTIVE   PROJECTION,   CONTINUED  121 

The  nearest  corner  of  the  cottage  is  at  a  certain  distance  into  the  picture, 
i.e.,  beyond  the  Picture  Plane,  and  also  at  a  distance  to  the  left  of  the  spectator, 
that  is,  to  the  left  of  the  central  line  of  vision,  or  Line  of  Sight. 

The  plan  of  the  cottage  is  arranged  so  that  the  length  way  of  it  is  at  an 
angle  of  40°  to  the  Picture  Plane.  At  a  convenient  place  to  the  left,  on  the  Ground 
Line  as  base,  is  erected  an  end  elevation,  from  which  levels  for  the  vertical 
traces  of  lines,  such  as  that  for  the  roof  ridge,  may  be  found,  from  which  to  draw 
vanishing  lines  to  the  vanishing  points.  These  vanishing  lines  on  the  Picture 
Plane,  as  was  seen  above,  are  intersections  of  the  Picture  Plane  by  oblique  planes, 
each  containing  the  eye  and  some  line,  such  as  the  ridge  line  of  the  roof,  and  there- 
fore containing  the  projection  of  it,  by  radial  projection  from  the  said  line  to 
the  eye.  In  Fig.  114,  the  drawing  has  been  made  as  economically  as  possible, 
in  order  to  save  confusion  of  lines.  It  will  be  readily  seen  that  by  continuing 
in  the  same  manner,  windows  and  door  ways,  etc.,  might  be  added. 

The  student  will  doubtless  realize  that,  if  it  were  necessary  to  find  vanishing 
points  for  oblique  Hnes,  these  would  be  found  in  the  vertical  traces,  on  the  Picture 
Plane,  of  vertical  planes  through  the  eye,  parallel  to  such  oblique  lines.  For  the 
purposes  of  our  present  study,  however,  sufficient  has  been  undertaken. 


EXERCISE  XLVI 

1.  Work  out  to  a  scale,  such  as  |"  to  the  ft.,  a  plan  and  elevation  for  a  structure  such  as  that 
given  in  the  Fig.  1 14,  where  the  sizes  are  15  ft.  by  26  ft.,  with  roof  plan  ig  ft.  by  30  ft.  The  nearest 
corner  of  the  roof  is  4  ft.  into  the  picture  and  3  ft,  to  the  left  of  the  spectator.  The  distance 
of  the  spectator  is  40  ft.,  that  is,  36  ft.  from  the  picture  plane.  Suitable  heights  and  details  should 
be  chosen.     Height  of  the  eye  6  ft. 

2.  Numerous  problems  will  readily  suggest  themselves,  such  as  books  lying  unevenly  upon 
one  another,  pieces  of  furniture,  buildings,  etc. 


iNDi:x 


Asymptotes.  70 
Attitude.  2 
Axes  of  projection.  71 
Axis,  of  cone,  34 

of  cylinder.  32 

of  rotation,  11 1 
Axometric  projection,  71 

Centre  of  vision,  (W.,  114 
Chords,  of  circles,  1 1 

of  the  sphere,  53 
Circles,  projected,  11 

as  great  circles  of  spheres,  04 
as  H.T.'s  of  right  vertical  cones,  83 
as  sections  of  right   cones  and   cylin- 
ders, 70 
as  sections  of  si)heres  with  spheres,  105 
shadows  of,  2^^ 
("omplement  angle,  14 
Compound  angle,  27 
Cone,  projection  of.  34 
sections  of,  7S 
shadow  of,  36 
tangent  planes  to.  S2,  8q 
I  races  of.  70 
Cones  enveloping  spheres,  g6.  97 
Cube,    projected    under    ditlerenl    conditions. 

I'igs.  28,  20.  30,  45.  66,  73,  Qi 
Curved  surfaces,  traces  of,  7q 
Cylinders,  projections  of.  ,^2 
sections  of.  77 
shadows  of.  37 
tangent  planes  to.  S5 
traces  of.  71) 


Develo{)ments  of  surfaces.  io-\  103 

Dihedral  angles.  56 

Distances  from  planes  of  projection,  4 

Elevations,  2 


Kllil)se,    the    foreshortened    a|jpearance    of    a 
circle,  1 1 
the  section  of  a  cylinder  or  cone,  77 
the  trace  of  a  cylinder  or  of  a  cone.  80 

Eye  level.  116 


Foreshortened  appearance.  5.  6 


Generator  lines.  7 
Ground  line.  1 14 


Helicoid.  1 1 1 

Helix.  1 1 1 

Horizon  line.  1 14 

Horizontal  diameter  of  inclined  circle 
lines  on  oblique  plane.  45 
plane  of  projection.  H.I*.. 

Hyperbola.  78 

Hyperboloid  of  revolution.  1 1 1 


Inclination,  5 

of  lines  to  oblique  planes.  44 

of  lines  to  planes  of  projection.  7 

of  planes  to  each  other,  60 

Interpenetrations  of  solids,  loi.  105.  io6,  laS 

Intersection,  of  lines  with  planes.  18.  4q 

of  planes  with  each  other,  48 

Isometric  projection,  71 


l.ine.  generator,  of  a  cylinder  or  cone.  77.  7g.  01 
of  intersection.  AT.  2 
of  intersection  of  oblique  planes.  48 
of  intersection  of  surfaces  of  interpcne 

trating  solids.  105.  107 
representing  ray  of  light.  20,  35.  1 14 
shadow  of.  :o 
tangent  to  great  cirrle  of  a  sphere.  o<) 


l.M 


124 


INDEX 


Lines,  as  edges  and  diagonals  of  figures,  4,  5        I    Projection  plane,  I 

at  inclinations  to  both  planes  of  pro-       Pyramids,  projection  of,  33,  65,  91 

jection,  27  j  sections  of,  76 

inclinations,  attitudes,  and  true  lengths  shadows  of,  35,  3S 

of,  6,  7 
inclined  to  oblique  planes,  44 
inclined  to  one  another  and  projected,       Rabattement,  of  great  circle  of  a  sphere,  94 

43,  62 
in  planes,  45 

perpendicular  to  planes,  46 
projection  lengths  of,  25 


Notation  adopted  for  naming  points  and  their 
projections,  etc.,  4 


Oblique  planes,  found,  angles  given,  28,  q6 
angles  contained  by,  57 
rabatted,  41,  42 
represented  by  lines,  i,  14 

Octahedron,  66 

Orthographical  or  perpendicular  projection,  i, 
114 


Parabola,  78 

Parallel  perspective,  118 

planes,  51 
Penetration,  see  Interpenet rations. 
Perspective  projection,  114 
Picture  plane,  114 
Pitch,  113 

IMane  figures,  projection  forms  of,  q,  10 
projection  of,  20 
shadows  of,  22 
Planes,  oblic}ue,  13 

angles  contained  by,  56 
intersection  of,  48 

parallel,  51 

rabattement  of,  40 
Plans,  2 
Points,  at  distances  from  planes  of  projection,  4 

on  oblique  planes,  45 

on  planes  of  projection,  17 

on  surface  of  sphere,  54,  03 

planes  passing  through,  53 

tangent,  on  spheres  by  planes,  gS 
Prisms,  32,  88 

development  of  surfaces  of,  102 

interpenetration  of,  101 


of  planes,  40 

of    right    angles    in    axometric 
projection,  74 

of  sections  of  solids,  76 

to    determine    angles    between 
lines,  42 

to  determine  dihedral  angles,  56 
Radial  projection,  114 
Rays  of  light,  20,  35,  114 
Revolution,  surfaces  of,  iii 


Scale,  73 

Screw-threads,  112 
Secondary  elevation,  of  a  figure,  11 
of  solids,  107 
Secondary  XY  line,  11,  107 
Sections  of  solids,  with  curved  surfaces,  77,  78 

with  plane  surfaces,  76 
Shade,  36 

Shadow  of  sphere,  81 
Shadows,  of  figures,  22 
of  lines,  20 
of  solids,  35 
Skeleton  cube,  71 
Solids,  in  perspective,  114,  120 

projection  of,  31,  88 
Sphere,  found,  when  points  on  its- surface  are 
given,  53 
great  circle  of,  94 
interpenetrating    with    other    spheres 

or  with  cones,  105 
shadow  of,  81 
tangent  planes  to,  93,  97 
to  find  points  on  surface  of,  93 
with  enveloping  cones,  96,  98 
Spiral,  1 1 1 

Supplement  angle,  56 
Surfaces  of  revolution,  iii 


Tangents  at  inlinity  to  the  hyperbola,  78 
Tangent  planes,  to  cones,  82 

to  cylinders,  8^ 


INDEX 


V2'i 


Tangent  planes,  to  spheres,  03 
Tetrahedron,  66 

height  of,  67 
Traces,  of  curved  surfaces,  79 
of  lines,  iS 
of  planes,  i,^ 
True  forms,  of  figures  by  rabattement.  47 

of  sections  of  solids.  76 
True  inclinations,  of  lines  with  each  other  by      XY  line,  2 
rabattement  of  plane  containing  them.  42  A'sKj.  or  secondary  AT  line,  11,  107 


!    True  inclinations,  of  lines  to  planes.  44 
True  lengths  of  lines,  7 

Vertical  plane  of  projection,  V.P..  i 
Vertical    transparent    picture    plane    in    per- 
sfK'Ctive  projection.  115 


14  DAY  USE 

KBTORN  TO  DBSK  BROM  WH.CH  BORKOWBO 

I  OAN  OEPT. 

Renewed  booUs  are  s,-— --d—»"- 


LD  21-100m-6,'56 
(B9311sl0)476 


General  Library     . 
Universiry  of  California 
Berkeley 


YD  05053 


-f\' 


UNIVERSITY  OF  CALIFORNIA  UBRARY 


